Solutions for Trigonometry by Gelfand & Saul

Git commit 6496780. Published July 8, 2024.

This work is licensed under a Creative Commons

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Introduction

Trigonometry by Gelfand and Saul is often recommended as a precalculus text

for self-study. However, those who are learning without the help of a teacher

can struggle with the lack of solutions to exercises in the text. Partial sets of

solutions for Trigonometry have been published by John Beach1and Fardeen

Ashraf2. It is hoped that this document will eventually contain a complete set of

solutions. Contributions are welcome. These can take the form of pull requests

or issues submitted to the project’s GitHub repository3.

Chapter 0: Trigonometry

Page 8

1. Statement I applies:

c2=a2+b2= 102+ 242= 100 + 576 = 676

c=√676 = 26

2. Statement I applies:

a2+b2=c2

a2+ 92= 412

a2+ 81 = 1681

a2= 1600

a=√1600 = 40

1https://jbeach50.weebly.com/gelfand–saul-trig-solutions.html

2https://archive.org/details/gelfand-trigonometry-solutions-manual

3https://github.com/philip-healy/gelfand-trigonometry-solutions

3. 52+ 122= 25 + 144 = 169 = 132. By Statement II, a right triangle exists

with legs of length 5 and 12, and hypotenuse of length 13.

4. Statement I applies:

a2+b2=c2

a2+ 12= 32

a2+ 1 = 9

a2= 8

a=√8 = √4√2=2√2

5. Statement I applies, where a=b:

a2+a2=c2

a2+a2= 12

2a2= 1

a2=1

a=r1

2=√1

√2=1

√2

6. From the diagram at the bottom of Page 11, we can see the shorter leg is

half the length of the hypotenuse. So in this instance the shorter leg has

length 1/2. We can use Statement 1 to ﬁnd the length of the longer leg:

a2+b2=c2

a2+1

22

= 12

a2+1

4= 1

a2=3

a=r3

4=√3

√4=√3

7. For any point Y, we can draw a triangle with sides AY ,BY and AB. Let

abe the length of side AY ,bbe the length of side BY and cbe the length

of side AB. According to Statement II, the subset of these triangles where

a2+b2=c2are right triangles with legs of length aand band hypotenuse

c. Let Xbe the subset of Ythat are vertices of these right triangles. This

set of points describes a circle with its centre at the midpoint of AB, and

radius AB/2.

Page 9

1. 62+ 82= 36 + 64 = 100 = 102. By Statement II on Page 7 (converse of the

Pythagorean Theorem), this is a right triangle.

2. 10-24-26 (Exercise 1), 9-40-41 (Exercise 2), 5-12-13 (Exercise 3)

3. Using the Pythagorean Theorem:

c2=a2+b2= 82+ 152= 64 + 225 = 289

c=√289 = 17

4. The ﬁrst column in the table increases by 3, the second increases by 4 and

the third increases by 5. Continuing to add rows yields triangles 12-16-20,

15-20-25 and 18-24-30.

5. Shortest side with length 10: 10-24-26. Shortest side with length 15: 15-36-

39.

6. Multiplying all sides by the common denominator (5), we get a similar tri-

angle with sides 15/5 = 3, 20/5 = 4 and 5. We know that this is a right

triangle from the table in Question 4.

7. To ﬁnd a similar triangle with shorter leg 1, divide all sides by 3, resulting in

sides 1-4/3-5/3. To ﬁnd a similar triangle with longer leg 1, divide all sides

by 4, resulting in sides 3/4-1-5/4.

8. To ﬁnd a similar triangle with hypotenuse 1, divide all sides by 13, resulting

in sides 5/13-12/13-1. To ﬁnd a similar triangle with shorter leg 1, divide

all sides by 5, resulting in sides 1-12/5-13/5. To ﬁnd a similar triangle with

longer leg 1, divide all sides by 12, resulting in sides 5/12-1-13/12.

9. To formula for the area of a triangle is 1

2bh where bis the length of the base

and his the height. For right triangles, ﬁnding the area is easy: one leg is the

base and the other leg is the height. For other triangles, ﬁnding the height

is more diﬃcult: we need to ﬁnd the length of the altitude drawn from the

base. The triangles with sides 5-12-13 and 9-12-15 are both right triangles:

see Exercise 3 on Page 8 and Exercise 4 on Page 9. The triangle with sides

13-14-15 is not a right triangle. We can conﬁrm this using Statement I:

a2+b2= 132+ 142= 365, c2= 152= 225, a2+b2=c2. However, if we join

the 5-12-13 and 9-12-15 triangles using their equal legs, the resulting triangle

has the dimensions we are looking for: 13-14-15. The base of this combined

triangle has length 5 + 9 = 14. We also know the length of the altitude from

the base of the combined triangle: 12. So, the area of the 13-14-15 triangle

is 1

2·14 ·12 = 84 units squared.

10. (a)

(b)

Page 11

1. 1

√2(see the solution for Question 5 on page 8).

Challenge: 1

√2=√2

2(multiplying above and below by √2). √2 is given to

9 decimal places in the diagram on the top of page 11: 1.4141213562373.

Dividing this decimal representation by 2 (using long division if necessary)

yields a ﬁgure of 0.707060678.

2. c2=a2+b2= 32+ 32= 9 + 9 = 18. c=√18 = √9√2 = 3√2.

3. The hypotenuse of a 30◦right triangle is double the length of the shorter leg.

In this instance the hypotenuse is 10 units long. We can use the Pythagorean

Theorem to ﬁnd the length of the longer leg:

a2+b2=c2

a2+ 52= 102

a2+ 25 = 100

a2= 75

a=√75 = √25√3=5√3

4. We can solve these by ﬁnding similar triangles to the 30◦right triangle with

sides 1-√3-2, or the 45◦right triangle with sides 1-1-√2.

(a) x=√3, y= 2

(b) x=1

√3,y=2

√3

(d) x= 4√3, y= 8

(e) x=y= 2√2

(f) x= 5, y= 5√2

Page 14 (Examples)

1. Why didn’t we need to compare 32with 22+ 42, or 22with 32+ 42?

The obtuse angle will always be opposite the longest side.

2. This conclusion is incorrect. Why?

From the footnote at the beginning of Chapter 0: “Given three arbitrary

lengths. . . they form a triangle if and only if the sum of any two of them is

greater than the third.” In this case 1 + 2 = 3 which is equal to (not greater

than) the third side.

Page 14 (Exercise)

1. (a) c2= 82= 64. a2+b2= 62+ 72= 36 + 49 = 85. c2< a2+b2, so the

triangle is acute.

(b) c2= 102= 100. a2+b2= 62+ 82= 36 + 64 = 100. c2=a2+b2, so

the triangle is a right triangle.

is acute.

(d) aand bare the same as in question b), but cis larger, so the triangle

is obtuse.

(e) c2= 122= 144. a2+b2= 52+ 122= 25 + 144 = 169. c2< a2+b2, so

the triangle is acute.

(f) c2= 142= 196. a2+b2= 169, as above. c2> a2+b2, so the triangle

is obtuse.

(g) The sum of two sides must be larger than the third, but 12 + 5 = 17

in this case.

Chapter 1: Trigonometric Ratios in a Triangle

Page 23

1. (a) sin α= 5/13

(b) sin α= 4/5

(d) c=√62+ 82=√100 = 10. sin α= 8/10.

(e) sin α= 3/5

(f) sin α= 12/13

(g) sin α= 3/5

(h) c=√72+ 32=√58. sin α= 7/√58.

2. (a) sin β= 12/13

(b) sin β= 3/5

(d) sin β= 6/10

(e) sin β= 4/5

(f) sin β= 5/13

(g) sin β= 4/5

(h) sin β= 3/√58

3. The example 30-60-90 triangle given on page 11 has sides 1, √3, 2. Let β

represent the 60◦angle. The opposite leg bhas length √3. The hypotenuse

chas length 2. So, sin β=b/c =√3/2≈1.732/2=0.866.

Crossing oﬀ the numbers listed:



0.1



0.2



0.3



0.4



0.5



0.6



0.7



0.8 0.9

Page 25

1. The Altitude-on-Hypotenuse Theorem tells us that when an altitude is drawn

to the hypotenuse of a right triangle, the two triangles formed are similar

to the given triangle and to each other. Therefore, the triangles with sides

a-b-c,a-p-dand d-b-qare similar, and the ratio for sin αappears in all of

them:

(a) b/c

(b) d/a

2. (a) sin α=h/b

(b) Multiplying both sides of formula above by b:h=bsin α

rewritten as: bc sin α/2.

(d) sin β=h/a. Rewriting this in terms of h:h=asin β. Substituting

this for hin the area formula: ac sin β/2.

(e) Let h2represent the altitude from Ato BC.sinβ =h2/c. Rewriting

in terms of h2, we get h2=csin β.

3. (a) Expressing hin terms of sin αand b:

sin α=h

h=bsin α

Expressing hin terms of sin βand a:

sin β=h

h=asin β

(b) Both expresssions are equal to h:

asin β=h=bsin α

sin β=h2

h2=csin β

Expressing h2in terms of sin γand b:

sin γ=h2

h2=bsin γ

Both expressions are equal to h2:

bsin α=h2=csin γ

(d) i. We can rewrite the result from part (b) so that the expressions on

each side are fractions with sine denominators:

asin β=bsin α

asin β

sin αsin β=bsin α

sin αsin β

sin α=b

sin β

ii. We can rewrite the result from part (c) similarly:

csin β=bsin γ

csin β

sin βsin γ=bsin γ

sin βsin γ

sin γ=b

sin β

We can derive the Law of Sines by combining results i. and ii. using the

common expression b/ sin β:

sin α=b

sin β=c

sin γ

Page 26

1. (a) cos α= 12/13. cos β= 5/13.

(b) cos α= 3/5. cos β= 4/5.

(d) cos α= 6/10. cos β= 8/10.

(e) cos α= 4/5. cos β= 3/5.

(f) cos α= 5/13. cos β= 12/13.

(g) cos α= 4/5. cos β= 3/5.

(h) cos α= 3/√58. cos β= 7/√58.

2. (a) c=√82+ 62=√64 + 36 = √100 = 10. cos α= 8/10. cos β= 6/10.

(b) c=√52+ 122=√25 + 144 = √169 = 13. cos α= 12/13. cos β=

5/13.

length of c. Next, we will use the Pythagorean Theorem to ﬁnd the

length of the longer leg:

a2+b2=c2

102+b2= 202

b2= 400 −100 = 300

b=√300 = √100√3 = 10√3

We can now ﬁnd cos αand cos β:

cos α=10√3

20 =√3

cos β=10

20 =1

(d) The triangle is congruent to the one above, so the solution is the same.

(e) Consider the 45◦right triangle with legs of length 1 and hypotenuse

√2. cos α= cos β= 1/√2.

(f) c=√32+ 42=√9 + 16 = √25 = 5. cos α= 3/5. cos β= 4/5.

(g) b=x√3. cos α=x√3/2x=√3/2. cos β=x/2x= 1/2.

3. The Altitude-on-Hypotenuse Theorem tells us that when an altitude is drawn

to the hypotenuse of a right triangle, the two triangles formed are similar

to the given triangle and to each other. Therefore, the triangles with sides

a-b-c,a-p-dand d-b-qare similar, and the ratio for cos αappears in all of

them:

(a) a/c

(b) p/a

Page 28

1. In this instance, α= 29◦,β= 61◦, and α+β= 90◦. According to the

theorem above, if α+β= 90◦, then sin α= cos β.

2. x= 90 −35 = 55◦

3. If α+β= 90◦, then β= 90◦−α. According to the theorem above, sin α=

cos β. Substituting (90 −α) for β: sin α= cos (90 −α).

Page 29

First, we need to ﬁnd the length of the hypotenuse: c=√32+ 42=√9 + 16 =

√25 = 5.

1. sin2α=4

52=16

2. sin2β=3

52=9

3. cos2α=3

52=9

25 (same as sin2β)

4. cos2β=4

52=16

25 (same as sin2α)

5. sin2α+ cos2α=16

25 +9

25 =25

25 = 1

6. sin2α+ cos2β=16

25 +16

25 =32

7. cos2α+ sin2β=9

25 +9

25 =18

Page 30

1. sin2α+ cos2α=4

52

+3

52

=16

25 +9

25 =25

25 = 1

2. It’s not an error. According to the corollary of the Pythagoream Theorem,

this a right triangle: a2+b2= 32+ 42= 9 + 16 = 25 = c2.

3. sin2β+ cos2β=3

52

+4

52

25 +16

25 =25

25 = 1

4. cos2α+ sin2α= 1

cos2α= 1 −sin2α= 1 −5

132

= 1 −25

169 =144

169

cos α=r144

169 =12

5. cos2α+ sin2α= 1

cos2α= 1 −sin2α= 1 −5

72

= 1 −25

49 =24

cos α=r24

49 =√4√6

√49 =2√6

6. We will follow the proof at the bottom of Page 29:

sin2α+ sin2β=a

c2

+b

c2

=a2

c2+b2

=a2+b2

a2+b2

= 1

7. Again, we will follow the proof at the bottom of Page 29:

cos2α+ cos2β=b

c2

+a

c2

=b2

c2+a2

=a2+b2

a2+b2

= 1

Page 31

angle xsin xcos x

30◦1

√3

45◦1

√2

60◦√3

α4

β3

2. cos 30◦=√3

2= sin 60◦

3. sin230◦+ cos230◦=1

22

+ √3

2!2

4+3

4= 1

4. We can observe from the table that sin xincreases with the size of an acute

angle (sin 30◦<sin 45◦<sin 60◦), while cos xdecreases with the size of an

acute angle. You can compare the fractions or convert to decimal make sure.

We know that sin α=4

5. We also know that αis an acute angle.

Is it larger or smaller than 30◦?Larger, 4

5>1

2so sin α > sin 30◦.

Than 45◦?Larger, 4

5>1

√2so sin α > sin 45◦.

Than 60◦?Smaller, 4

5<√3

2so sin α < sin 60◦.

Page 33 (First)

1. As the angle αget smaller, the ratio of the opposite side to the hypotenuse

approaches 0.

2. Recall from the theorem on page 28 that if α+β= 90◦, then sin α= cos β

and cos α= sin β. So, if sin 90◦= 1, then cos 0◦= 1.

3. sin20◦+ cos20◦= 02+ 12= 0 + 1 = 1

4. sin290◦+ cos290◦= 12+ 02= 1 + 0 = 1

5. Our friend is mistaken; the sine of an angle can never be greater than 1.

Page 33 (Second)

sin 0◦+ cos 0◦0 + 1 1

sin 30◦+ cos 30◦1

2+√3

21.366 (approx.)

sin 45◦+ cos 45◦1

√2+1

√21.414 (approx.)

sin 60◦+ cos 60◦√3

2+1

21.366 (approx.)

sin 90◦+ cos 90◦1 + 0 1

sin α+ cos α, where α

is the smaller. . .

5+4

51.4

sin α+ cos α, where α

is the larger. . .

5+3

51.4

2. If sin α= 1, then cos α= 0 and sin α+ cos α= 1. If cos α= 1, then

sin α= 0 and sin α+ cos α= 1. Otherwise, sin α < 1 and cos α < 1, so

sin α+ cos α < 2.

3. First we will expand and simplify (sin α+ cos α)2:

(sin α+ cos α)2= sin2α+ 2 sin αcos α+ cos2α

= (sin2α+ cos2α) + 2 sin αcos α

= 1 + 2 sin αcos α

We know that 0 ≤sin α≤1 and 0 ≤cos α≤1 because αis acute. So

2 sin αcos αis the product of three nonnegative numbers, and is itself a

nonnegative number. A nonnegative number added to 1 results in a number

≥1. Therefore, 1+2 sin αcos α≥1. The square root of a number ≥1 is itself

≥1. Therefore, √1 + 2 sin αcos α≥1. Rewriting the expression on the left:

√1 + 2 sin αcos α=p(sin α+ cos α)2= sin α+ cos α. So, sin α+ cos α≥1.

4. sin 45◦+ cos 45◦=1

√2+1

√2=2

√2=2√2

√2√2=2√2

2=√2

5. You should notice that the values for sin α+cos αincreases with larger alpha

when 0◦≤α < 45◦, reaches a maximum value when α= 45◦, then decreases

with larger αwhen 45◦< α ≤90◦.

Page 35

(sin 0◦)(cos 0◦) 0 ·1 0

(sin 30◦)(cos 30◦)1

2·√3

20.433 (approx.)

(sin 45◦)(cos 45◦)1

√2·1

√20.5

(sin 60◦)(cos 60◦)√3

2·1

20.433 (approx.)

(sin α)(cos α), where α

is the smaller. . .

5·4

50.48

(sin α)(cos α), where α

is the larger. . .

5·3

50.48

How large can the product (sin α)(cos α)get? We can see from the table that

the maximum value of the product appears to be when α= 45◦.

Page 37

1. cos α= 3/5, cos β= 4/5, sin α= 4/5, sin β= 3/5, tan α= 4/3, tan β= 3/4,

cot α= 3/4, cot β= 4/3.

2. We can show that this assumption is correct using the corollary of the

Pythagorean Theorem: a2+b2= 32+ 42= 25 = c2.

3. cos α=a/c, cos β=b/c, sin α=b/c, sin β=a/c, tan α=b/a, tan β=a/b,

cot α=a/b, cot β=b/a.

4. c=√122+ 52=√169 = 13. cos α= 12/13. cos β= 5/13. cot α= 12/5.

cot β= 5/12.

5. First, we will use the Pythagorean Theorem to ﬁnd the length of the longer

leg:

a2+b2=c2

a2+ 72= 252

a2+ 49 = 625

a2= 576

a= 24

We can now ﬁnd the numerical values that were asked for: cos α= 24/25,

cos β= 7/25, cot α= 24/7, cot β= 7/24.

6. a

c= sin α= cos β

c= cos α= sin β

b= tan α= cot β

a= cot α= tan β

7. First, we will use the Pythagorean Theorem to ﬁnd the length of the other

leg:

a2+b2=c2

a2+ 32= 52

a2+ 9 = 25

a2= 16

a= 4

We can now ﬁnd the numerical values that were asked for: cos α= 4/5,

cot α= 4/3.

8. If tan α= 1, then a/b = 1, implying that a=band α= 45◦. cos α=

cos 45◦= 1/√2. cot α= 1/1 = 1.

9. tan 45◦= 1/1 = 1.

10. tan 30◦= 1/√3≈0.57735.

11. tan 45◦+ sin 30◦= 1 + 1

2=3

2. We don’t need a calculator because both

numbers are rational.

Chapter 2: Relations among Trigonometric Ra-

tios

Page 43

1. cos α=s1−8

172

=r1−64

289 =r225

289 =15

tan α=

cot α=15

2. Let the length of the adjacent leg abe 3

7and the length of the hypotenuse

be 1 (see the ﬁrst triangle diagram on page 44).

sin α=p1−a2=s1−3

72

=r1−9

49 =r40

49 =√4√10

√49 =2√10

tan α=√1−a2

2√10

=2√10

cot α=a

√1−a2=3

2√10

3. sin α=√1−b2,tan α=√1−b2

b,cot α=b

√1−b2

4. sin α=d

√1 + d2,cos α=1

√1 + d2,cot α=1

sin αcos αtan αcot α

sin α a √1−a2a

√1−a2

cos α√1−a2a√1−a2

√1−a2

tan αa

√1 + a2

√1 + a2a1

cot α1

√1 + a2

Page 45 (First)

1. Given in text

2. sin245◦=1

√22

sin αcos αtan αcot α

sin αsin αp1−sin2αa

p1−sin2αp1−sin2α

sin α

cos α√1−cos2αcos α√1−cos2α

cos α

√1−cos2α

tan αtan α

√1 + tan2α

√1 + tan2αtan α1

tan α

cot α1

√1 + cot2α

cot α

√1 + cot2α

cot αcot α

Page 45 (Second)

1. tan α=a

b= cot β

2. cot α=b

a= tan β

3. sec α=c

a= csc β

4. csc α=c

b= sec β

Page 47

1. (a) sin230◦+ cos230◦=1

22

+ √3

2!2

4+3

4= 1

(b) sin245◦+ cos245◦=1

√22

+1

√22

2+1

2= 1

2!2

+1

22

4+1

4= 1

2. sin2α+ cos2α= 1

√5

4!2

+ cos2α= 1

cos2α= 1 − √5

4!2

= 1 −5

16 =11

cos α=r11

16 =√11

3. sin2α+ cos2α= 1

sin2α+2

32

= 1

sin2α= 1 −4

9=5

sin α=r5

9=√5

4. sin α

cos α= tan α=1

√3

sin2α

cos2α=1

sin2α

1−sin2α=1

3 sin2α= 1 −sin2α

4 sin2α= 1

sin2α=1

sin α=r1

4=1

sin α

cos α= tan α=1

√3

sin2α

cos2α=1

1−cos2α

cos2α=1

3∗1−cos2α

cos2α= 3 ∗1

3∗(1 −cos2α)

cos2α= 1

3∗(1 −cos2α) = cos2α

3−3 cos2α= cos2α

3 = 4 cos2α

4= cos2α

√3

2= cos α

And then to check our solution we can calculate the fraction we are given

√3from our cos αand sin αfractions.

√3

2∗√3

√3

5. (a) cot xsin x=1

tan xsin x=sin x

tan x=sin x

sin x

cos x

=sin xcos x

sin x= cos x

(b) tan x

sin x=

sin x

cos x

sin x=

sin x

cos x·1

sin x

sin x·1

sin x

sin xcos x

1=sin x

sin xcos x=1

cos x

(d) This one is tricky. You might need to try a few diﬀerent approaches

(squaring above and below, multiplying above and below by cos αsin α).

Eventually it becomes clear that you need to multiply above and below

by (1 −cos α) and ﬁnd a way to cancel out the sin αfactor in the

numerator:

sin α

1 + cos α=sin α(1 −cos α)

(1 + cos α)(1 −cos α)=sin α(1 −cos α)

1−cos α+ cos α−cos2α

=sin α(1 −cos α)

1−cos2α=sin α(1 −cos α)

1−(1 −sin2α)=sin α(1 −cos α)

1−1 + sin2α

=sin α(1 −cos α)

sin2α=1−cos α

sin α

(e) sin2α+ 2 cos2α−1

cot2α=1−cos2α+ 2 cos2α−1

cot2α=cos2α

cos α

sin α2

=cos2α

cos2α

sin2α

=cos2αsin2α

cos2α

= sin2α

(f) cos2α=cos2α

1=cos2α

cos2α+ sin2α

cos2α

cos2α+sinα

cos2α

cos2α+sin2α

cos2α

1 + tan2α

(g) sin2α=sin2α

1=cos2α

cos2α+ sin2α

sin2α

cos2α+sinα

sin2α

cos2α

sin2α+sin2α

sin2α

cot2α+ 1

(h) 1−cos α

1 + cos α=(1 −cos α)(1 + cos α)

(1 + cos α)(1 + cos α)=1 + cos α−cos α−cos2α

(1 + cos α)2

=1−cos2α

(1 + cos α)2=sin2α

(1 + cos α)2

=sin α

1 + cos α2

(i) The key to solving this one is the formula for factoring a diﬀerence of

cubes: a3−b3= (a−b)(a2+ab +b2).

sin3α−cos3α

sin α−cos α=(sin α−cos α)(sin2α+ sin αcos α+ cos2α)

sin α−cos α

= sin2α+ sin αcos α+ cos2α

= 1 + sin αcos α

6. (a) We can rewrite the LHS to show that sin4α−cos4α= cos2α−sin2α:

sin4α−cos4α= (sin2α+ cos2α)(sin2α−cos2α) = 1(sin2α−cos2α)

= sin2α−cos2α

Answer: There are no angles αfor which sin4α−cos4α > cos2α−sin2α

because the expressions on either side of the inequality are equivalent.

(b) sin4α−cos4α >= cos2α−sin2αfor all angles αbecause the expressions

on either side of the inequality are equivalent.

7. If we rewrite 2 sin αcos αas a fraction, we can divide above and below by

cos αto convert the numerator and denominator into expressions in terms

of tan α:

2 sin αcos α=2 sin αcos α

1=2 sin αcos α

sin2α+ cos2α

2 sin αcos α

cos2α

sin2α+cos2α

cos2α

2 sin α

cos α

sin2α

cos2α+cos2α

cos2α

=2 tan α

tan2α+ 1

Now we can plug in the given value for tan αto ﬁnd the value of 2 sin αcos α

in this instance:

2 sin αcos α=2 tan α

tan2α+ 1 =2( 2

5)2+ 1 =

25 + 1 =

25 +25

=20

8. First, we will rewrite the expresssion cos2α−sin2αin terms of tan α:

cos2α−sin2α=cos2α−sin2α

1=cos2α−sin2α

cos2α+ sin2α=

cos2α−sin2α

cos2α

cos2α+sin2α

cos2α

=1−tan2α

1 + tan2α

(a) To ﬁnd the numerical value of cos2α−sin2αwhen tan α=2

5we can

substitute 2

5for tan αin the formula above:

cos2α−sin2α=1−tan2α

1 + tan2α=1−2

52

1 + 2

52=1−4

1 + 4

=21

(b) Substituting rfor tan αin the formula above:

cos2α−sin2α=1−r2

1 + r2

9. First, we will rewrite the expresssion in terms of tan α:

sin α−2 cos α

cos α−3 sin α=

sin α−2 cos α

cos α

cos α−3 sin α

cos α

sin α

cos α−2 cos α

cos α

cos α−3 sin α

cos α

=tan α−2

1−3 tan α

Next, we substitute 2

5for tan α:

tan α−2

1−3 tan α=

5−2

1−32

5=

5−10

5−6

=−8

−1

= 8

10. First, we will rewrite the expresssion in terms of tan α:

asin α+bcos α

ccos α+dsin α=

asin α

cos α+bcos α

cos α

ccos α

cos α+dcos α

cos α

=atan α+b

c+dtan α

Next, we substitute 2

5for tan αand simplify:

atan α+b

c+dtan α=a2

5+b5

5

c5

5+d2

5=

2a+5b

5c+2d

=2a+ 5b

5c+ 2d

Now we can see why the problem included the restriction that 5c+ 2d= 0;

the value of the expression is undeﬁned if the denominator is zero. The sum

of two rational numbers is a rational number. Therefore the numerator and

denominator in the expression are both rational numbers. The quotient of

two rational numbers is a rational number. Therefore, the entire expression

evaluates to a rational number for arbitratrary rational values of a,b,cand

11. We can expand and simplify the expression:

(sin α+ cos α)2+ (sin α−cos α)2

= sin2α+ 2 sin αcos α+ cos2α+ sin2α−2 sin αcos α+ cos2α

= 2 sin2α+ 2 cos2α

= 2(sin2α+ cos2α)

= 2(1)

= 2

As the expression evaluates to a constant, it is as large as possible for all

values of α.

Page 49

1. Rewriting any instances of sec αor csc αon either side of the identities:

(a) tan αcsc α= sec α

tan α1

sin α=1

cos α

tan α

sin α=1

cos α

(b) cot αcsc α= sec α

cot α1

cos α=1

sin α

cot α

cos α=1

sin α

sec αcsc α= cot α

cos α·1

sin α= cot α

cos α1

sin α= cot α

cos α

sin α= cot α

(d) tan2α= (sec α+ 1)(sec α−1)

tan2α= sec2α−1

tan2α=1

cos2α−1

(e) csc2α= 1 + cot2α

sin2α= 1 + cot2α

2. Rewriting any instances of sin αor cos αon either side of the identities, and

eliminating fractions:

(a) tan α

sin α=1

cos α

tan α1

sin α= sec α

tan αcsc α= sec α

(b) 1

sin αcos α= cot α

cos α

sin α= cot α

cot α= cot α

cos2α

tan2α+ 1 = sec2α

(d) 1

sin2α= 1 + cot2α

csc2α= 1 + cot2α

Page 50

1. First, we ﬁnd the value of a2+b2:

a2+b2= (cos2α−sin2α)2+ (2 sin αcos α)2

= cos4α−2 cos2αsin2α+ sin4α+ 4 sin2αcos2α

= cos4α+ 2 cos2αsin2α+ sin4α

= (cos2α+ sin2α)2

= (1)2

= 1

According to the lemma on Page 50, as a2+b2= 1, an angle θexists such

that a= cos θand b= sin θ.

2. First, we ﬁnd the value of a2+b2:

a2+b2= r1 + cos α

2!2

+ r1−cos α

2!2

=1 + cos α

2+1−cos α

=1 + cos α+ 1 −cos α

= 1

3. First, we will rewrite aand bto eliminate the cube exponents:

a= 4 cos3α−3 cos α

= 4 cos αcos2α−3 cos α

= 4 cos α(1 −sin2α)−3 cos α

= 4 cos α−4 sin2αcos α−3 cos α

= cos α−4 sin2αcos α

b= 3 sin α−4 sin3α

= 3 sin α−4 sin αsin2α

= 3 sin α−4 sin α(1 −cos2α)

=−sin α+ 4 sin αcos2α

Next, we will expand a2and b2:

a2= (cos α−4 sin2αcos α)2

= cos2α−8 sin2αcos2α+ 16 sin4αcos2α

b2= (−sin α+ 4 sin αcos2α)2

= sin2α−8 sin2αcos2α+ 16 sin2αcos4α

Next, we add the expressions for a2and b2and simplify to 1:

a2+b2= cos2α−8 sin2αcos2α+ 16 sin4αcos2α+ sin2α−8 sin2αcos2α+

16 sin2αcos4α

= cos2α+ sin2α−16 sin2αcos2α+ 16 sin4αcos2α+ 16 sin2αcos4α

= cos2α+ sin2α+ 16 sin2αcos2α(−1 + sin2α+ cos2α)

= 1 + 16 sin2αcos2α(0)

= 1

According to the lemma on Page 50, as a2+b2= 1, an angle θexists such

that a= cos θand b= sin θ.

4. First, we ﬁnd the value of a2+b2:

a2+b2=1−t2

1 + t22

+2t

1 + t22

=(1 −t2)2

(1 + t2)2+(2t)2

(1 + t2)2

=(1 −t2)2+ (2t)2

(1 + t2)2

=1−2t2+t4+ 4t2

(1 + t2)(1 + t2)

=(1 + t2)(1 + t2)

(1 + t2)(1 + t2)

= 1

According to the lemma on Page 50, as a2+b2= 1, an angle θexists such

that a= cos θand b= sin θ.

5. We expand (p2−q2)2+ (2pq)2and use the fact that p2+q2= 1 to simplify

to 1:

(p2−q2)2+ (2pq)2=p4−2p2q2+q4+ 4p2q2

=p4+ 2p2q2+q4

= (p2+q2)2

= (1)2

= 1

This is similar to Exercise 1 above.

Page 51

1. sin α < 1 when αis acute, therefore 1−sin α > 0 when αis acute. 1−sin α=

0 when sin α= 1, i.e., α= 90◦.

2. cos α < 1 when αis acute, therefore 1−cos α > 0 when αis acute. 1−cos α=

0 when cos α= 1, i.e., α= 0◦.

3. Statement a) is always true. Statements b) and c) both include the case that

sin2α+ cos2α= 1, which is always true.

4. Let xbe the maximum cost of the items in a supermarket. In Supermarket

A, x≤$1. In Supermarket B, x < $1. In Supermarket C, x≤$1. In

Supermarket D, x > $1. We can see that Supermarkets A and C are oﬀering

the same terms.

5. Inequality a) is correct. For b) to be correct, an angle αwould have to

exist such that sin α+ cos α= 2. We know that this is not the case. When

α= 90◦, sin α= 1 and cos α= 0. When α= 0◦, sin α= 0 and cos α= 1.

When 0◦< α < 90◦, sin α < 1 and cos α < 1. In all cases, sin α+ cos α < 2.

6. The largest possible value of sin αis 1, and occurs when α= 90◦. The largest

possible value of cos αis 1, and occurs when α= 0◦. See Page 32.

Page 52

1. sin 30◦= 0.5, sin 45◦= 0.707, sin 60◦= 0.866.

2. By using the tan button to calculate tan 60◦, and the sqrt button to calcu-

late √3, Betty can compare the results: both are 1.732.

3. Press tan, then enter the angle degree measure, then press 1/x

in radical or rational form

αsin αcos αtan αcot α

30◦1

√3

√3√3

45◦1

√2

√21 1

60◦√3

2√31

√3

in decimal form, from calculator

αsin αcos αtan αcot α

30◦0.5 0.866 0.577 1.732

45◦0.707 0.707 1 1

60◦0.866 0.5 1.732 0.577

Page 53

1. The sine of the larger angle is 4/5 = .8. We can use the inverse sine function

to ﬁnd the angle: arcsin .8 = 53.1301◦. The sum of the three angles in the

triangles is: arcsin .6 + arcsin .8 + 90◦= 36.8699◦+ 53.1301◦+ 90◦= 180◦.

2. (a) arcsin 1 = 90◦

(b) arccos 0.7071067811865 = 45◦

3. arccos 0.8 = 36.8699◦

4. arcsin 0.6 = 36.8699◦

5. Half of sin 30◦(0.25) seems like a reasonable estimate. The actual value is

0.2588.

6. Entering arcsin 0.3 in calculator, we get 17.458◦

If we enter sin 17.458◦

7. With arcsin x= 53◦, to ﬁnd value of x, we do x= sin 53◦, which gives us

x= 0.7986

If we take arcsin 0.7986, we will get 52.9966

8. x=√3/2

9. arcsin(sin 17◦) = 17◦

10. sin(arcsin 0.4) = 0.4

11. arcsin(sin 30◦)) = 30◦

arcsin 1/2 = 30{circ}. Since sin 30◦= 0.5 as in question 5

12. cos220◦+ sin220◦= 1

cos280◦+ sin280◦= 1

13. tan 20◦= 0.36397 and sin 20◦

cos 20◦=0.34202

0.93969 ≈0.36397

Note tan 80◦= 5.67128 and sin 80◦

cos 80◦=0.98481

0.17365 ≈5.67128

14.

2π3π

22π

−1

Page 55

α= 72

To ﬁnd a

sin α=a

sin 72◦=a

a= 9 ×sin 72◦

a≈8.5595

To ﬁnd b

cos α=b

cos 72◦=b

b= 9 ×cos 72◦

b≈2.7812

To ﬁnd the hypotenuse c, use pythagoras theorem

c=pa2+b2

c=√149

To ﬁnd α

tan α=b

tan α=7

α= arctan 7

α≈34.992◦

To ﬁnd β

tan β=a

tan β=10

β= arctan 10

β≈55.008◦

β= 27

To other acute angle will be 90 −27 = 63 degrees

To ﬁnd length of hypotenuse c

sin β=12

c=12

sin 27

= 26.432

To ﬁnd length of the other leg a

tan β=12

a=12

tan 27

= 23.551

α= 73

To other acute angle will be 90 −73 = 17 degrees

To ﬁnd length of hypotenuse c

sin β=20

c=20

sin 73

= 20.914

To ﬁnd length of the other leg a

tan β=20

a=20

tan 73

= 6.115

Page 56

α= 46

To ﬁnd length of shadow a

tan α=50

a=50

tan 46

a= 48.284

Therefore the length of shadow is 48.284 feet

α= 32

To ﬁnd length of shadow a

tan α=a

a= 35 tan 32

a= 21.87

Therefore the height of ﬂagpole is 21.87 feet

3. Shadow is the longest in the morning as the sun is at lower altitude

Shadow is shortest during mid day as the sun is exactly above the pole

4. There will be no shadow when the sun is exactly above the ﬂag pole, which

happens at a speciﬁc time during ’zero shadow day’ (search online for exact

time and day for your location)

Page 59

1. Draw point C on the left side of circle, by circle geometry, the angle in that

triangle vertex will be θ/2. Then apply the relation sin α=AB

2r, where AB

is the opposite side of angle α. Therefore sin θ/2 = AB since radius is 1/2

2. Draw point C on the left side of circle, by circle geometry, the angle in that

triangle vertex will be θ/2. Then apply the relation sin α=AB

2r, where AB

is the opposite side of angle α. Therefore sin θ/2 = AB

2since radius is 1,

rearranging the equation gives 2 sin θ/2 = AB

3. From question 1, for circle of diameter 1, we have the relation sin θ/2 = AB.

Therefore if angle ϕ>θthen the opposite side of angle ϕwill be longer too,

therefore sin ϕ > sin θ

4. Assuming the three points are on a circle, and realising that the sides 6, 8

and 10 form a right angle triangle, therefore the side of length 10 would have

been the diameter of the circle. Therefore the radius for the circle is 5

5. Since sin α=P B

2r, re-arranging the equation gives P B

sin α= 2r. That is, the

ratio of a: sin αis equal to the diameter of circle (double of radius)

6. let a be opposite side of angle αlet b be opposite side of angle βlet c be

opposite side of angle γ

Exercise shows that a

sin α= 2r,b

sin β= 2r,c

sin γ= 2r. Since all are equal to

2rthen they are ratios with same value, that is

sin α=b

β=c

7. Since sin α=AB

2×1/2, then chord AB = sin α. Since arc AC is double of arc

AB, then chord AC = 2 sin α

But the diagram shows that in triangle ABC and using triangle inequality,

AB +BC ≥AC. Therefore sin 2α < 2 sin α

8. Using sin α=AB

2r, if α= 60◦and radius is 10/2=5

The chord AB would be 2 ×5×sin 60◦= 5√3 units long

9. Using sin α=AB

2r, if α= 30◦(as central angle is double of angle on circle)

and radius is 10/2=5

The chord AB would be 2 ×5×sin 30◦= 5 units long

10. The diagonal of a square inside a circle will be the diameter of circle and the

angle of each triangle corner is 45◦

Using sin α=AB

2r, where 2r is 10 (diameter) and angle α= 45◦, we get

AB = 10 sin 45◦= 5√2 units long for each side of square inside a circle

11. Using sin α=AB

2r, then letting one side of pentagon AB = 2r×sin α.

Nothing that a pentagon can be cut into three distinct triangles, therefore

the internal angles add up to 180 ×3 = 540 making each internal angle

540 ÷5 = 108 degrees large

As each vertex is divided into three equal angles, it will give α= 108÷3 = 36

degrees. As diameter is 10, each side of pentagon will be 10 ×sin 36

Page 62

1. The degree measure of a semicircle is 180◦. The degree measure of a quarter

circle is 90◦.

2. The measure of arc cut oﬀ by one side of regular pentagon inscribed in a

circle is 360◦/5 = 72◦. For a regular hexagon: 360◦/6 = 60◦. For a regular

octagon: 360◦/8 = 45◦.

Page 64

1. Let AB be an arc on the circle and points Q and R are two other points on

the circle

The angle in the centre will be double of angle AQB, which will be double

of angle ARB as well, therefore the both aangles AQB and ARB must be

equal

2. A regular pentagon can be divided into three equal circles, therefore a pen-

tagon’s total internal angle is 180 ×3 = 540 degrees.

Since the intenal angles are equal size for a regular pentagon, then each angle

is 540 ÷5 = 108 degrees large

3. Let points ABCD be points on a circle, then draw line AC. Note that the

angle AOC will be double of ABC

Meawhile on the other side of AOC will be double of ADC, these two

sides makes a full revolution (360 degrees). Therefore we have 2ABC +

2ADC = 360, that is ABC +ADC = 180 degrees

Page 65

1. As it is larger than 60 degrees, then the points will be inside the circle

2. As it is 90 degrees, then the points will be on the circle with AB being the

diameter of the circle

3. Triangle AQB would have a larger inscribed circle of radius R, therefore the

relation sin α=AB

2Rthat is 2R=AB

sin α

Triangle APB is inside a smaller circle of radius r, therefore the relation

sin Q=AB

2r. That is 2r=AB

sin Q

The ratio r:R will be AB

sin Q÷AB

sin α=sin α

sin Q. That is r

R=sin α

sin Q

if r < R then sin α < sin Q. That is α < Q for the left diagram if r > R

then sin α > sin Q. That is α > Q for the right diagram

4. Following above proof, let angle Q be 0.5α. Since sin α > sin 0.5α, then

r > R, so it will be inscribed in the circle of original arc

5. On the same side, the angle subtended will be α, where angle subtended in

the centre would be 2α. The other side of the centre angle would be 360−2α,

which gives 180 −αas size of subtended angle on the other arc side

Note, this is proof for opposite angles of a cyclic quadrilateral are supple-

mentary

Chapter 3: Relationships in a Triangle

Page 68

Data Determine a triangle? Restrictions?

1ABa Yes 0◦< mA, mB, mA+mB < 180◦

2ABb Yes (Duplicate of 1) 0◦< mA, mB, mA+mB < 180◦

3ABc Yes 0◦< mA, mB, mA+mB < 180◦

4AbC Yes (Duplicate of 3) 0◦< mA, mB, mA+mB < 180◦

5ABC No One side length must also be given.

6Abc Yes 0◦< mA < 180◦

7Bbc No b≥csin B. Also, mAmust be given in some cases.

8Cbc No (Duplicate of 7) c≥bsin C. Also, mAmust be given in some cases.

Page 70

B C

hc=bsin α

= 12 sin 70◦

≈11.28

3. Let us consider a right triangle ABC with a right angle at A.

B CP

βγ

cha

Analyzing the right triangle AP B in the above diagram gives us that ha=

csin β. Similarly, analyzing the right triangle AP C in the above diagram

gives us that ha=bsin γ. This is consistent with our formulas for ha.

B C

βγ

b=hc

c=hb

Since Ais a right angle, band care themselves altitudes, as shown by the

diagram above. Since the sine of a right angle is equal to 1, this is consistent

with the formulas hb=csin αand hc=bsin α. Additionally, by analyzing

the sines at angles Band C, we ﬁnd that hb=asin γand hc=asin β,

which again agrees with the formulas for hband hc.

R Q

30◦

A=1

2php

2pq sin R

2(12) (10) sin 30◦

2(12) (10) 1

2

= 30

Page 71

β γ

For the hbformulas, there is nothing new to check as αand γare both acute

angles.

From the above diagram, since △ARC is a right triangle, we can see that

hc=bsin α. Also, since △BRC is a right triangle, we have that hc=

asin (180◦−β). Since we have asserted that the sine of an angle should be

equal to the sine of its supplement, the previous formula is equivalent to

hc=asin β.

Page 73

1. We ﬁrst substitute awith b,bwith c,cwith a,αwith β,βwith γ,γwith

α.a

sin α=b

sin β→b

sin β=c

sin γ

We now substitute awith c,cwith b,bwith a,αwith γ,γwith β,βwith

α.a

sin α=b

sin β→c

sin γ=a

sin α

Combining these equalities together gives the Law of Sines:

sin α=b

sin β=c

sin γ.

2. Let A= 30◦,B= 60◦,C= 90◦. We know that b=a√3 and c= 2aby the

properties of 30-60-90 triangles.

sin α=a

1/2= 2a

sin β=a√3

√3/2= 2a

sin γ=2a

1= 2a

Since all of the above ratios are equal to 2a, we have shown that the Law of

Sines holds for 30-60-90 triangles.

3. In the triangle on the left, the angle opposite the side of length 10 measures

60◦. If xis the side length opposite the 50◦angle and yis the side length

opposite the 70◦angle, then

sin 50◦=10

sin 60◦=⇒x=10 sin 50◦

sin 60◦≈8.85

sin 70◦=10

sin 60◦=⇒y=10 sin 70◦

sin 60◦≈10.85

In the triangle on the right, the missing angle measures 60◦. If xis the side

length opposite the 60◦angle and yis the side length opposite the 65◦angle,

then x

sin 60◦=12

sin 55◦=⇒x=12 sin 60◦

sin 55◦≈12.69

sin 65◦=12

sin 55◦=⇒y=12 sin 65◦

sin 55◦≈13.28

4. As discussed in Section 4 of this chapter, the formulas for the altitudes using

the sine function remain unchanged for obtuse triangles if we assert that the

sine of an angle is equal to the sine of its supplement. Therefore, the proof of

the Law of Sines for obtuse triangles is unchanged from the proof presented

in the beginning of this section.

5. In the triangle on the left, the angle opposite the side of length 6 measures

45◦. If xis the side length opposite the 15◦angle and yis the side length

opposite the 120◦angle, then

sin 15◦=6

sin 45◦=⇒x=6 sin 15◦

sin 45◦≈2.20

sin 120◦=6

sin 45◦=⇒y=6 sin 120◦

sin 45◦≈7.35

In the triangle on the right, the missing angle measures 115◦. If xis the

side length opposite the 40◦angle and yis the side length opposite the 115◦

angle, then

sin 40◦=14

sin 25◦=⇒x=14 sin 40◦

sin 25◦≈21.29

sin 115◦=14

sin 25◦=⇒y=14 sin 115◦

sin 25◦≈30.02

6. We ﬁrst solve for the angle opposite the side of length 10.

sin 50◦=10

sin α=⇒sin α=10

8sin 50◦= 0.9576 =⇒α= 73.2◦

We now solve for the third angle by subtracting the other two angles from

180◦.

β= 180◦−50◦−73.2◦= 56.8◦

7. Since the sine of an angle is equal to the sine of its supplement, based on

our calculations in the previous exercise, αcould equal 73.2◦or 106.8◦. This

implies that βcan equal 56.8◦or 23.2◦. The below diagram shows the two

triangles that satisfy the given constraints overlaid on top of each other.

50◦

106.8◦

73.2◦

8. This result follows immediately from the theorem proven in Section 12 of

Chapter 2.

Page 75 (First)

2R=10

sin 30◦=⇒R=10

2 sin 30◦=10

2(1/2) = 10

30◦

60◦

Alternatively, we can ﬁnd the radius from ﬁrst principles. Let AP B be a

triangle inscribed in a circle Owith AB = 10 and mAP B = 30◦. Since

AP B is an inscribed angle, the corresponding central angle AOB has a

measure of 60◦. We also know that segments OA and OB are congruent

because they are both radii of the circle O. Therefore, △AOB is isosceles.

This combined with the fact that the measure of AOB is 60◦implies that

△AOB is equilateral. Because AB is equal to 10, we know that the the

radius of the circle is also 10 because the other two sides of the equilateral

triangle are radii.

2R=8

sin 90◦=⇒R=8

2 sin 90◦=8

2(1) = 4

Alternatively, we can apply Thales’s theorem (see the appendix of Chapter

2) to show that the hypotenuse of an inscribed triangle is a diameter of the

circumscribing circle. If the diameter of the circle is 8, then the radius would

be 4.

Page 75 (Second)

A=1

2(8) (11) sin 40◦= 44 sin 40◦≈28.3

2. (a)

A=1

2(10) (9) sin 23◦= 45 sin 23◦≈17.6

(b)

A=1

2(3) (7) sin 130◦= 10.5 sin 130◦≈8.0

(c)

A=1

2(3) (7) sin 90◦= 10.5 sin 90◦= 10.5

Since this is a right triangle, the sides adjacent to the 90◦angle are

altitudes, so we can use the more elementary formula for the area of a

triangle: A=1

2bh.

3. Let bbe the length of AC.

40 = 1

2b(6) sin 40◦=⇒b=40

3 sin 40◦≈20.7

4. Let αbe the measure of angle P.

9 = 1

2(5) (6) sin α=⇒sin α=9

15 =⇒α≈36.9◦or α≈143.1◦

5. The area of this triangle is 1

2ab sin γ, where γis the angle included between

the sides of length aand b. Because sin γhas a maximum value of 1 when

γis 90◦, the area of the triangle is maximized when γis 90◦. This gives an

area of 1

2ab. Since γ= 90◦, the triangle is a right triangle.

In this triangle, aand bare both legs of the right triangle. We can let one

of aor bbe the hypotenuse of a right triangle to get another non-congruent

right triangle with side lengths of aand b. Without loss of generality, we

can choose bto be the hypotenuse. By the Pythagorean theorem, the other

leg of the triangle has length √b2−a2. Therefore, the area of this triangle

is 1

2apb2−a2.

6. As in the previous exercise, for two given side lengths of a triangle, the area

of the triangle is maximized when the angle included between these two sides

is equal to 90◦. Thus, the maximum area of an isosceles triangle with a leg

length of xis 1

2(x) (x) sin 90◦=1

2x2.

A B

CD a

b b

Because opposite angles of a parallelogram are congruent, the area of the

above parallelogram is given by

2ab sin A+1

2ab sin C=1

2ab sin C+1

2ab sin C=ab sin C.

Notice that Aand Care congruent, Band Care supplements, and Cand

Dare supplements by the properties of parallelograms. This means that the

above area formula can be used with any angle of the parallelogram since

sin A= sin B= sin C= sin D.

180◦−α

A=1

2wx sin α+1

2xy sin (180◦−α) + 1

2yz sin α+1

2wz sin (180◦−α)

2wx sin α+1

2xy sin α+1

2yz sin α+1

2wz sin α

2(wx +xy +yz +wz) sin α

2(w(x+z) + y(x+z)) sin α

2(w+y) (x+z) sin α

w+yand x+zare the lengths of the diagonals in the quadrilateral in the

diagram above. The above derivation shows that the choice of αdoes not

matter since sin α= sin (180◦−α).

180◦−α

A=1

2(d+z)xsin α+1

2(d+z)ysin (180◦−α)−1

2xz sin α−1

2yz sin (180◦−α)

2(d+z)xsin α+1

2(d+z)ysin α−1

2xz sin α−1

2yz sin α

2[(d+z)x+ (d+z)y−xz −yz] sin α

2[(d+z) (x+y)−z(x+y)] sin α

2d(x+y) sin α

10. 1

2(4) (6) sin α

2(8) (10) sin α

=4·6

8·10 =3

11. See solution provided in the textbook

12.

180◦−α

|AP B|×|CP D|=1

2wx sin α1

2yz sin α=1

4wxyz sin2α

|BP C|×|DP A|=1

2xy sin (180◦−α)1

2wz sin (180◦−α)=1

4wxyz sin2α

This identity is true even if the diagonals intersect outside of the quadrilat-

eral. Using the diagram above in Exercise 9, we can show that |AP B| ×

|CP D|and |BP C|×|DP A|are both equal to

4xyz (d+z) sin2α.

13.

c=AB =AP +P B =bcos A+acos B

Let’s suppose angle Ais obtuse. Then we need to subtract . . .

c=AB =AP −BP =bcos A−acos (180◦−B) = bcos A+acos B

As shown above, no changes need to be made to this result for obtuse angles.

Page 79

x=acos B′

b2=h2+ (c+x)2

=a2−x2+c2+ 2cx +x2

=a2+c2+ 2cx

=a2+c2+ 2ac cos B′

Page 80

2. (a)

x2=y2+z2−2yz cos X

(b)

x2=y2+z2−2yz cos X

(c)

a2=b2+c2−2bc cos A

(d)

b2=a2+c2−2ac cos B

(e)

c2=a2+b2−2ac cos C

3. Given two sides and the angle between them, the Law of Cosines can be

used to ﬁnd the length of the third side. Another application of the Law of

Cosines can be used to solve for one of the unknown angles since the three

sides of the triangle are now known. The ﬁnal angle of the triangle can be

determined by subtracting the measures of the two known angles from 180◦.

4. (a)

x=p122+ 152−2 (12) (15) cos 50◦≈11.7

(b)

x=p102+ 162−2 (10) (16) cos 110◦≈21.6

(c)

cos x◦=62+ 82−92

2 (6) (8) =19

96 =⇒x≈78.6

(d)

cos x◦=102+ 52−122

2 (10) (5) =−19

100 =⇒x≈101.0

cos A=102+ 72−62

2 (10) (7) =113

140 =⇒A≈36.2◦

cos B=102+ 62−72

2 (10) (6) =29

40 =⇒B≈43.5◦

cos C=72+ 62−102

2 (7) (6) =−5

28 =⇒C≈100.3◦

A B

CD 12

3 3

In the above diagram of a parallelogram with side lengths of 3 and 12, let

d1be the length of the diagonal AC and let d2be the length of the diagonal

BD. By the Law of Cosines, we know that

1= 32+ 122−2(3)(12) cos B

and

2= 32+ 122−2(3)(12) cos A.

However, since adjacent angles in a parallelogram are supplementary, cos B=

−cos A. Therefore,

1+d2

2=32+ 122−2(3)(12) cos B+32+ 122−2(3)(12) cos A

=32+ 122+ 2(3)(12) cos A+32+ 122−2(3)(12) cos A

= 32+ 122+ 32+ 122

= 306

Thus, the answer does not depend on the speciﬁc shape of the parallelogram.

Only the side lengths are pertinent.

A B

CD a

b b

We can generalize the argument from the previous exercise to an arbitrary

parallelogram. As before, let d1be the length of the diagonal AC and let d2

be the length of the diagonal BD. Then,

1+d2

2=a2+b2−2ab cos B+a2+b2−2ab cos A

=a2+b2+ 2ab cos A+a2+b2−2ab cos A

=a2+b2+a2+b2,

which shows that the sum of the squares of the diagonals of a parallelogram

is equal to the sum of the squares of the four sides.

B A

We can create a parallelogram from triangle ABC by creating a copy of

the triangle and reﬂecting it through the point M. This allows us to apply

the result from the previous exercise about the sum of the squares of the

diagonals.

AD2+BC2=AB2+AC2+BD2+CD2=⇒(2AM )2+BC2=AB2+AC2+AC2+AB2

=⇒4AM2+BC2= 2AB2+ 2AC2

=⇒4AM2= 2AB2+ 2AC2−BC2

9. We can get a formula for the length of each median in a triangle by applying

cyclic substitutions to the result we derived in the previous exercise.

4AM2= 2AB2+ 2AC2−BC2

4BM2= 2BC2+ 2AB2−AC2

4CM2= 2AC2+ 2BC2−AB2

Adding these three equations together gives

4AM2+ 4BM2+ 4CM2= 3AB2+ 3AC2+ 3BC2.

Factoring out 3 on the right-hand side of the equation and dividing both

sides by 4 gives the result:

AM2+BM2+CM2=3

4AB2+AC2+BC2.

10.

11.

c2= 12+ 42−2 (1) (4) cos 60◦= 17 −81

2= 13 =⇒c=√13

12.

c2=a2+b2−2ab cos 60◦=a2+b2−2ab 1

2=a2+b2−ab

Since 120◦is the supplement of 60◦, the analogous result is that c2=a2+

b2+ab.

13.

Let A,B, and Cbe the positions of the three riders at some time t, where t

is the number of elapsed hours since departure. Then AX = 60t,BX = 40t,

and CX = 20t. Applying the result from the previous exercise concerning

the Law of Cosines for triangles with a 120◦angle,

AB =q(60t)2+ (40t)2+ (60t) (40t)≈87.2t

BC =q(40t)2+ (20t)2+ (40t) (20t)≈52.9t

AC =q(60t)2+ (20t)2+ (60t) (20t)≈72.1t

Therefore, after 1 hour, riders Aand Bare 87.2 miles apart, riders Band

Care 52.9 miles apart, and riders Aand Care 72.1 miles apart. After 2

hours, riders Aand Bare 174.4 miles apart, riders Band Care 105.8 miles

apart, and riders Aand Care 144.2 miles apart.

Page 83

1. 45-45-90 triangle:

2. The triangles which can be placed on their mirror images without reﬂection

are the isosceles triangles.

3. Rectangle:

Rhombus:

Isosceles Trapezoid:

Page 84

sin β=2S

ac =bc

ac =b

sin γ=2S

ab =bc

ab =c

c2=a2+b2−2ab cos γ

=a2+b2−22S

sin γcos γ

=a2+b2−4Scot γ

Page 85

b2=a2+c2−2ac cos β

a2=b2+c2−2bc cos α

2. These three quantities are all equal to twice the circumradius of the triangle.

Page 86

1. (a) The dimensions of the left-hand side and right-hand side of the equation

are both [L]/[∅], where we use ∅to represent a dimensionless quantity.

(b) The dimensions of the left-hand side are [L]/[L], which is a dimension-

less quantity. The dimensions of the right-hand side are [∅]/[∅], which

is also a dimensionless quantity. Thus, the dimensions in this equation

are consistent.

right-hand side are [L]·[L]·[∅], which is equivalent to the dimensions

of [L]2.

(d) The dimensions of the left-hand side are [L]2. The dimensions of each

term on the right-hand side are [L]2(recall that cos γis dimensionless),

so the overall dimensions of the right-hand side are also [L]2.

Page 88

1. The semiperimeter of this triangle is (3 + 4 + 5)/2 = 6.

S=p6 (6 −3) (6 −4) (6 −5) = 6

2. The semiperimeter of this triangle is (5 + 12 + 13)/2 = 15.

S=p15 (15 −5) (15 −12) (15 −13) = 30

3. The semiperimeter of an equilateral triangle with side length lis 3l/2

S=r3l

2·l

2=r3

16l4=l2√3

4. Each angle in an equilateral triangle is 60◦.

S=1

2l2sin 60◦=l2√3

5. The semiperimeter of a 13-14-15 triangle is (13 + 14 + 15)/2 = 21.

S=p21 (21 −13) (21 −14) (21 −15) = 84

The semiperimeter of a 25-39-56 triangle is (25 + 39 + 56)/2 = 60.

S=p60 (60 −25) (60 −39) (60 −56) = 420

The semiperimeter of a 25-39-16 triangle is (25 + 39 + 16)/2 = 40.

S=p40 (40 −25) (40 −39) (40 −16) = 120

Chapter 4: Angles and Rotations

Page 93

1. (a)

160◦

(b)

190◦

(c)

400◦

(d)

600◦

(e)

1200◦

(f)

−70◦

(g)

−400◦

(h)

360◦

(i)

−270◦

Page 98

1. (a)

30◦

Because 390◦= 360◦+ 30◦, a point rotated through an angle of 390◦

would end up at the same location as a point rotated through an angle

of 30◦. Therefore,

sin 390◦= sin 30◦=1

(b)

120◦

60◦

Because 3720◦= 10 ·360◦+ 120◦, a point rotated through an angle of

3720◦would end up at the same location as a point rotated through an

angle of 120◦. Therefore,

cos 3720◦= cos 120◦=−cos 60◦=−1

(c)

45◦

Because 1845◦= 5 ·360◦+ 45◦, a point rotated through an angle of

1845◦would end up at the same location as a point rotated through an

angle of 45◦. Therefore,

tan 1845◦= tan 45◦=sin 45◦

cos 45◦=√2/2

√2/2= 1.

(d)

315◦

45◦

Because 315◦= 360◦−45◦, a point rotated through an angle of 315◦

would end up at the same location as a point rotated through an angle

of −45◦. Therefore,

sin 315◦= sin (−45◦) = −sin 45◦=−√2

(e)

60◦

Because 420◦= 360◦+ 60◦, a point rotated through an angle of 420◦

would end up at the same location as a point rotated through an angle

of 60◦. Therefore,

cot 420◦= cot 60◦=cos 60◦

sin 60◦=1/2

√3/2=1

√3.

(f)

−30◦

tan (−30◦) = sin (−30◦)

cos (−30◦)=−sin 30◦

cos 30◦=−1/2

√3/2=−1

√3

2. (a)

tan 360◦= tan 0◦=sin 0◦

cos 0◦=0

1= 0

(b)

sin 180◦= sin 0◦= 0

(c)

cos 180◦=−cos 0◦=−1

(d)

cot 90◦=cos 90◦

sin 90◦=0

1= 0

(e)

cot 360◦= cot 0◦=cos 0◦

sin 0◦=1

0=⇒cot 360◦is undeﬁned.

(f)

tan (−270◦) = tan 90◦=sin 90◦

cos 90◦=1

0=⇒tan (−270◦) is undeﬁned.

Page 100

1. 400◦= 360◦+ 40◦, so Pwill lie in the ﬁrst quadrant.

3600◦= 10 ·360◦+ 0◦, so Pwill lie in the ﬁrst quadrant (on the positive

x-axis).

1845◦= 5 ·360◦+ 45◦, so Pwill lie in the ﬁrst quadrant.

−30◦=−360◦+ 330◦, so Pwill lie in the fourth quadrant.

−359◦=−360◦+ 1◦, so Pwill lie in the ﬁrst quadrant.

sin 30◦1/2 sin (−30◦)−1/2

sin 135◦√2/2 sin (−135◦)−√2/2

sin 210◦−1/2 sin (−210◦) 1/2

sin 300◦−√3/2 sin (−300◦)√3/2

sin 390◦1/2 sin (−390◦)−1/2

sin 480◦√3/2 sin (−480◦)−√3/2

From the table, we can see that sin (−α) = −sin α.

3. (a)

sin α= 0 =⇒α= 180◦

(b)

cos α= 0 =⇒α= 90◦,270◦

(c)

sin α= 1 =⇒α= 90◦

(d) cos α= 1 is true for α= 0◦and α= 360◦, but these are not in the

interval 0 < α < 360◦.

(e)

sin α=−1 =⇒α= 270◦

(f)

cos α=1

2=⇒α= 60◦,300◦

(g)

sin α=−1

2=⇒α= 210◦,330◦

(h)

sin2α=1

2=⇒sin α=±√2

2=⇒α= 45◦,135◦,225◦,315◦

(i) This equation has no solutions because the square of a real number

cannot be negative.

4. (a) If sin α= 5/13, then αcan lie in either the ﬁrst or second quadrant.

cos α= 12/13 when αis in the ﬁrst quadrant, and cos α=−12/13

when αis in the second quadrant.

(b) If sin α=−5/13, then αcan lie in either the third or fourth quadrant.

cos α= 12/13 when αis in the fourth quadrant, and cos α=−12/13

when αis in the third quadrant.

P= (a, b)

Let P= (a, b) be a point on the coordinate plane. Since a2+b2= 1, P

lies on a circle of radius 1 centered at the origin (the unit circle). Consider

an angle θbetween the positive x-axis and the line segment connecting P

to the origin. By the extended deﬁnitions of the sine and cosine functions,

sin θ=band cos θ=a.

Page 102

1. (a) Even

f(−x)=(−x)6−(−x)2+ 7 = x6−x2+ 7 = f(x)

(b) Odd

f(−x)=(−x)3−sin (−x) = −x3+ sin x=−f(x)

f(−x) = 1

−x+ 1

(d) Even

f(−x) = sec (−x) = 1

cos (−x)=1

cos x= sec x=f(x)

(e) Odd

f(−x) = csc (−x) = 1

sin (−x)=−1

sin (x)=−csc x=−f(x)

(f) Odd

f(−x) = 2 sin (−x) cos (−x) = −2 sin (x) cos (x) = −f(x)

(g) Even

f(−x) = sin2(−x)=(−sin x)2= sin2x=f(x)

(h) Even

f(−x) = cos2(−x) = (cos x)2= cos2x=f(x)

(i) Even

f(−x) = sin2(−x) + cos2(−x) = sin2x+ cos2x=f(x)

g(−x) = 1

2[f(−x) + f(−(−x))] = 1

2[f(−x) + f(x)] = g(x),

so g(x) is even.

h(−x) = 1

2[f(−x)−f(−(−x))] = 1

2[f(−x)−f(x)] = −h(x),

so h(x) is odd.

For any function f(x),

f(x) = 1

2[f(x) + f(−x)] + 1

2[f(x)−f(−x)] = g(x) + h(x),

so f(x) can be written as the sum of an even function and an odd function.

3. Following the notation of the previous question, we denote the even part of

f(x) as g(x) and the odd part of f(x) as h(x).

(a) Since we know cos xis an even function and sin xis an odd function,

g(x) = cos x, h (x) = sin x

(b) Since a polynomial is an even function when all variables are raised to

even exponents and an odd function when all variables are raised to

odd exponents,

g(x) = x2+ 1, h (x) = x3+x

g(x) = 1

22x+ 2−x, h (x) = 1

22x−2−x

(d) Applying the result from the previous exercise,

g(x) = 1

21−sin x

1 + sin x+1 + sin x

1−sin x=1

22 + 2 sin2x

1−sin2x=1 + sin2x

1−sin2x

h(x) = 1

21−sin x

1 + sin x−1 + sin x

1−sin x=1

2−4 sin x

1−sin2x=−2 sin x

1−sin2x

(e) Applying the result from the previous exercise,

g(x) = 1

21

x+ 2 +1

−x+ 2=1

24

4−x2=2

4−x2

h(x) = 1

21

x+ 2 −1

−x+ 2=1

2−2x

4−x2=−x

4−x2

Chapter 5: Radian Measure

Page 107

1. 180 degrees is equal to πradians. 90 degrees is equal to π/2 radians.

πr= 180◦=⇒1r= (180/π)◦=⇒2r= (360/π)◦≈114.6◦

3. Since a full rotation is 2πradians, 1/4 of a full rotation will be π/2 radians.

4. Since 45 degrees is 1/8 of a full rotation, 1/8 of a full rotation will be π/4

radians.

5. Filled table below

Degree Measure Radian Measure

90 π/2

180 π

270 3π/2

360 2π

90 π/2

180 π

270 3π/2

360 2π

6. Filled tables below

Degree Measure Radian Measure

30 π/6

72 2π/5

120 2π/3

135 3π/4

30 π/6

36 π/5

45 π/4

60 π/3

120 2π/3

126 7π/10

Degree Measure Radian Measure

198 11π/10

210 7π/6

216 6π/5

225 5π/4

240 4π/3

198 11π/10

200 10π/9

210 7π/6

216 6π/5

225 5π/4

240 4π/3

7. Since 360 degrees is equal to 2πradians, then 1 degree is equal to 2π/360,

or π/180, radians.

8. (a) sin(1r)≈0.8415

(b) sin(1◦)≈0.0175

9. Filled table below

α(in radian) sin αcos α

π/6 1/2√3/2

π/3√3/2 1/2

π/2 1 0

2π/3√3/2−1/2

7π/6−1/2−√3/2

5π/4−√2/2−√2/2

3π/2−1 0

11π/6−1/2√3/2

10. With an angle of 2 radians and a radius of 1, the length of the arc is 2×1 = 2.

With an angle of 3 radians and a radius of 1, the length of the arc is 3×1 = 3.

With an angle of πradians and a radius of 1, the length of the arc is π×1 = π.

11. With an angle of 2 radians and a radius of 3, the length of the arc is 2×3 = 6.

With an angle of 3 radians and a radius of 3, the length of the arc is 3×3 = 9.

With an angle of πradians and a radius of 3, the length of the arc is π×3 =

3π.

12. Using the fact that sine and cosine are cofunctions (see Section 4 of Chapter

1),

sin π

9= cos π

2−π

9= cos 7π

18 ,

so α= 7π/18.

13. Again, applying the properties of cofunctions, we know that sin α= cos (π/2−α).

14. A complete rotation around a circle corresponds to an angle of 2πradians,

so each of the six sectors in the diagram is an angle of 2π/6, which is a bit

more than 1 radian (since 2π > 6).

However, 2π/6 radians is equal to 60 degrees, so if 2π/6 is a bit more than

1, then 1 radian is less than 60 degrees

Geometric solution:

E F

In the diagram above, we place the six points Athrough Fso that they are

equally spaced across the circle O. This means that adjacent points on the

circle are separated by 360◦/6 = 60◦. Focusing on △AOB, we notice that

this triangle is isosceles because AO and BO are both equal to the radius

of the circle. This implies that OAB is congruent to OBA. Because the

three angles in a triangle total to 180◦, we have that

mAOB =mOAB =mOBA = 60◦.

In other words, △AOB is an equilateral triangle. This means that the side

AB is equal to the radius of O. Therefore, the arc that is intercepted by the

chord AB must be longer than the radius of Obecause the shortest path

between two points is a straight line. This implies that the central angle

AOB is greater than 1 radian. Since AOB is a 60◦angle, we have shown

that 60◦is greater than 1 radian.

Page 111

1. Since a circle of radius 1 traveling 1 foot corresponds to a rotation of 1

radian, a circle of radius 1 traveling 5 feet down a road corresponds to a

rotation of 5 radians.

2. From the previous question, the circle has rotated 5 radians, which is equal

to 5 ·180◦/π ≈286.48◦(two decimal places).

3. Since a circle of radius 1 rotates 1 radian to travel 1 foot, if it rotates 4

radians, then it has traveled 4 feet.

4. Since 120◦= 2π/3 radians, the wheel has traveled 2π/3 feet down the road.

5. A sector of radius 1 with angle αradians has an arc length of 1 ×α, so:

An angle of 1/2 radian has an arc length of 1/2.

An angle of π/2 radian has an arc length of π/2.

An angle of αradian has an arc length of α.

6. Since 2πradians equal to 360◦then

720◦= 4πradians

1440◦= 8πradians

3600◦= 20πradians

15120◦= 84πradians

12πradians = 2160◦

15πradians = 2700◦

100πradians = 18000◦

7. A sector of radius 3 with angle αradians has an arc length of 3α, so:

An angle of 1/2 radian has an arc length of 3/2.

An angle of π/2 radian has an arc length of 3π/2.

An angle of αradian has an arc length of 3α.

8. A sector of radius 3 and angle αhas arc length of 3α.

An angle of 1.5 radians has an arc length of 4.5.

9. A sector of radius 5 and angle αhas arc length of 5α.

An angle of 80 degrees is equal to 4π/9 radians. Thereforem its arc length

is 5 ×4π/9 = 20π/9.

10. A sector of radius 2 and angle αhas an arc length of 2α. This means that

for an arc length of α, its angle would be α/2. If the arc length is 3, then

its central angle is 3/2 radians.

11. A sector of radius 6 and angle αhas an arc length of 6α. This means that

for arc length of α, its angle would be α/6. If the arc length is 2, then its

central angle is 1/3 radians, which is equal to (60/π)◦≈19.1◦.

12. A circle of radius 7 units will travel 7 units with a rotation of 1 radian. That

is, each unit of travel requires a rotation of 1/7 radians. Therefore, 20 units

of travel require 20/7 radians of rotation.

13. A circle of radius 8 units will travel 8 units with a rotation of 1 radian. Since

150 degrees is equal to 5π/6 radians, the circle has rolled 8 ×5π/6 = 20π/3

units.

14. In twelve hours, the hour hand makes a complete rotation around the watch

face (e.g., 12:00 AM to 12:00 PM), so in one hour, the hour hand makes

1/12 of a full rotation. Because the hands of a clock rotate in a clockwise

sense, the angle of rotation will be negative. Thus, the angle through which

the hour hand rotates in one hour is

−1

12 ·2π=−π

15. In one hour, the minute hand makes one complete rotation around the watch

face, so it rotates through an angle of −2π. In the same time, the second

hand makes 60 complete rotations around the watch face since it completes

one rotation in one minute. Therefore, the second hand rotates through an

angle of −60 ·2π=−120π.

16. Because the hands of a watch travel clockwise, Joe’s angle should be negative

since counterclockwise rotations are positive by convention.

17. 1000 in.

2πin. ≈159.2

This trip consists of about 159 full rotations of the hour hand. Multiplying

by 12 gives the number of hours the trip takes since a complete rotation of

the hour hand takes 12 hours.

1000

2π·12 hours ≈1910 hours

18. The angular speed of the hour hand on a pocket watch and the hour hand on

Big Ben should be the same because in one hour, the hour hand travels the

same angular distance on both clocks. Since in both this exercise and the

previous exercise the clocks are traveling through an angle of 1000 radians,

both trips should take the same length of time: 1910 hours.

19. The wheel turns 2πradians before the spoke returns to the same position.

After the wheel turns πradians, the spoke will go from pointing vertically

downwards to pointing vertically upwards.

20. (a) A wheel of radius one meter will roll 2πmeters each revolution, so the

blue marks will be 2πmeters apart.

(b) The wheel will make a full revolution in this time, so it has rolled

through an angle of 2πradians.

coinciding. When they next coincide, the distance between the two

coincidences will be equal to 3n(measuring using the gaps between red

dots) or 2πm (measuring using the gaps between blue dots), where n

and mare positive integers. Because these are two equivalent ways to

measure the same distance, we can say that 3n= 2πm. Solving for π,

we ﬁnd that π=3n

2m. However, this is not possible as πis irrational,

so it cannot be expressed as the ratio of integers. This contradiction

means that if a red dot and blue dot do coincide, then they cannot

coincide again.

(d)

Let’s consider how the red marks hit the wheel as it rolls. Hitting three

red marks in a row corresponds to traveling 6 meters (count the space in

between the marks). Because the radius of the wheel is 1 meter, when

the wheel hits a triple of red marks, the last mark will hit the wheel in

a position that is 6 radians ahead of the position where the ﬁrst mark

was hit. Alternatively, we can say that the last mark hits the wheel in

a position that is 2π−6 radians behind the position of the ﬁrst mark.

In the above diagram, the red marks are separated by 2π−6 radians,

so they represent the positions of every other red mark that hits the

wheel in some period of time. Notice that the position of the red marks

on the wheel implies that the blue mark cannot be more than 2π−6

radians away from a red mark. This means that no matter how far the

wheel travels, the blue mark will eventually be placed within 2π−6

meters of a red mark.

Another perspective:

Compare the location of marks by plotting the marks’ locations

5 10 15 20 25 30 35 40

Note the distance between blue and red markers will be diﬀerent but

the closest they come together are

Red Blue Diﬀerence (approx)

6 2π0.283

12 4π-0.566

18 6π-0.850

24 8π-1.133

30 10π-1.416

(e) We assume that the blue dot and red dot coincide to begin with. 100

rotations corresponds to the wheel travelling 200πmeters. Notice that

200π≈209 ·3+1.3. Because 1

4·3<1.3<1

2·3, the blue dot will be

between two pink dots: one of the pink dots will be at the midpoint of

two red dots, and the second pink dot will be directly behind the ﬁrst.

Page 114

1. The calculator was in radian mode, since sin(1r)≈0.8415, while sin(1◦)≈

0.0175.

x(in radians) sin xDiﬀerence (approx.)

0.2 0.19867 1.33 ×10−3

0.15 0.14944 5.62 ×10−4

0.05 0.04998 2.08 ×10−5

In all cases, x > sin x.

3. In the below table, the diﬀerence being calculated is x−x3/6−sin x.

x(in radians) sin x x −x3/6 Diﬀerence (approx.)

0.2 0.1986693 0.1986667 −2.66 ×10−6

0.15 0.1494381 0.1494375 −6.32 ×10−7

0.05 0.0499792 0.0499792 −2.60 ×10−9

4. sin 10◦≈0.174, and 10/60 ≈0.167, so the error in the approximation is

0.007.

5. (a)

sin 0.1≈0.1−0.13

6=1

10 −1

6000 =599

6000

Error: sin 0.1 is greater than 599/6000 by 8.33 ×10−8.

(b)

sin 0.1◦= sin π

1800 ≈π

1800 ≈3.14

1800 =157

90000

Error: sin 0.1◦is greater than 157/90000 by 8.84 ×10−7.

6. (a) sin 1000◦≈ −0.9848

(b) sin 1000r≈0.8269

7. (a) sin(sin 1000r)≈0.7358

(b) sin 3.14r≈0.00159

8. Let ε=π/2−1.5707 ≈9.6×10−5.

cos 1.5707 = cos π

2−ε= sin ε < ε < 0.0001

Page 116

1. Using the graph below

−1π1π2π

−1

7π/5−3π/7

(a) sin 7π/5 is negative because π < 7π/5<2π.

Estimated x≈ −1 (actual is −0.9511)

(b) sin −3π/7 is negative because −π < −3π/7<0.

Estimated x≈ −1 (actual is −0.9749)

sin 13π/6 = 1/2. In general, sin x= 1/2 when x=π/6 + 2πn or

x= 5π/6+2πn, where nis an integer.

(d) Draw a line at y= sin(π/12) ≈0.26 and then estimate the location of

the intersections with the sine wave.

x=π/12,11π/12,25π/12, . . .

≈0.26,2.88,6.54, . . .

−1π1π2π

−1

0.26 2.88

(e) Draw a line at y= 0.8 and then estimate the location of the intersec-

tions with the sine wave. x≈0.93,2.21,7.21, . . .

−1π1π2π

−1

0.93 2.21

Alternatively, notice that sin π/4 = √2/2≈0.70 and sin π/3 = √3/2≈

0.85. Taking a weighted average, we can estimate that the sine of

3·π

4+2

3·π

3=11π

is approximately 0.8. Then, based on the properties of the sine function,

25π/36 and 83π/36 should also have sines of approximately 0.8.

Note: For parts (a) and (b) of the above exercise, we can use the approx-

imation cos x≈1−1

2x2to get better estimates. This approximation can be

derived from the approximation sin x≈xusing the identity sin2x+ cos2x= 1.

Let us suppose that xis a positive angle close to zero. Replacing sin xwith

xin the aforementioned identity, we get cos x≈√1−x2. Using the binomial

approximation, (1 + z)α≈1 + αz, we arrive at the desired approximation for

cos x:

cos x≈p1−x2=1 + −x21/2≈1−1

2x2.

For part (a), we get the estimate

sin 7π

5=−sin 2π

5=−sin π

2−π

10=−cos π

10 ≈1

2π

102

−1≈1

20−1 = −0.95.

For part (b), we get the estimate

sin −3π

7=−sin 3π

7=−sin π

2−π

14=−cos π

14 ≈1

2π

142

−1≈1

40−1 = −0.975.

Both of these estimates improve upon the ﬁrst-order approximation of −1.

Page 120

1. Results are written to 5 decimal places.

α(radians) α(degrees) sin α

1 57.29578 0.84147

0.5 28.64789 0.47943

0.2 11.45916 0.19867

0.1 5.72958 0.09983

0.01 0.57296 0.01000

0.02 1.14592 0.02000

0.001 0.05730 0.00100

0.002 0.11459 0.00200

0.005 0.28648 0.00500

2. Since sin x≈xfor angles close to zero (when xis in radians), sin 0.00123456 ≈

0.00123456. Since x > sin xfor positive angles, we know this is an overesti-

mate. The calculator result is 0.0012345597, so our estimate is accurate to

7 decimal places.

3. (a) Results are written to 5 decimal places.

α α −α3/6 sin α

1 0.83333 0.84147

0.5 0.47917 0.47943

0.2 0.19867 0.19867

0.1 0.09983 0.09983

0.05 0.04998 0.04998

0.01 0.00999 0.00999

0.001 0.00099 0.00099

(b) Recall that we can multiply the degree measure of an angle by π/180

to get the radian measure, so we simply replace αby πD/180 in our

approximation. This gives

sin D≈πD

180 −1

6πD

1803

=πD

180 −π3D3

34992000

For D= 1◦, the above approximation gives sin 1◦≈0.017421, while the

calculator result is 0.017452. This is an underestimate by 3.1×10−5.

4. Using the fact that α=πD/180 from the previous part, the largest possible

error for an angle measured in degrees is given by

120 πD

1805

=π5D5

22674816000000.

x5/120 <0.001

x5<0.12

x < 5

√0.12

x < 0.65438

6. The signs are alternating between terms, so the third term should be positive.

The exponent of xincreases by two in each successive term, so the third

term should have an exponent of 5.

The denominator of ﬁrst term is 1! and the denominator of second term is

3!. Therefore the denominator of the ﬁfth term should be 5! = 120.

Therefore, the third term is

+x5

120.

Hint: This is the Taylor Series for sin x.

Note: !is used to denote the factorial function, where the factorial of a

positive integer nis the product of all positive integers less than or equal to

7. The aliens appear to take clockwise angles to be positive, which is the op-

posite of the convention we typically use. Furthermore, it is not clear that

the aliens even use counterclockwise angles since all of the examples only

feature clockwise angles. This could mean that the aliens only work with

non-negative (or in our system, non-positive) angles.

The aliens do consider angles corresponding to rotations greater than a full

rotation, as we do. φseems to be used to denote one complete rotation.

Comparing to our system, φ=−2πradians.

8. It is easier to rewrite each of the radian angle measurements as multiples of

πsince the ﬁrst quadrant contains angles between 0 and 0.5π, the second

quadrant contains angles between 0.5πand π, etc.

Angle in radians Angle in terms of πQuadrant

1 0.318π1

2 0.636π2

3 0.954π2

4 1.273π3

5 1.591π4

6 1.909π4

1000 radians is approximately equal to 318.31π, which represents 159 full

rotations with a remainder of 0.31πradians. Thus an angle of 1000 radians

lies in the ﬁrst quadrant.

1000 degrees corresponds to 1000/360 ≈2.78 rotations, which represents 2

whole rotations and a further 0.78 of a rotation, which is slightly more than

a 3/4 rotation. Thus, an angle of 1000 degrees lies in the fourth quadrant.

9. Approximately 1/4 of the angles lie in each quadrant. See the solution below

the question in the textbook.

Chapter 6: The Addition Formulas

Page 123

α β sin αsin βsin α+ sin βsin (α+β)

60◦30◦√3/2 1/2√3+1/2 sin 90◦= 1

π/4π/4√2/2√2/2√2 + √2/2 = √2 sin π/2=1

π/6π/3 1/2√3/21 + √3/2 sin π/2=1

2. For these values of αand β, sin αand sin βare both at least 1/2. Further-

more, at least one of sin αand sin βis strictly greater than 1/2. Therefore,

sin α+ sin β > 1

2+1

2= 1 = sin (α+β).

3. (a)

sin 60◦+ sin 30◦=√3

2+1

sin (60◦+ 30◦) = sin 90◦= 1

This identity is not correct.

(b)

sin (60◦−30◦) = sin 30◦=1

sin 60◦−sin 30◦=√3

2−1

This identity is not correct.

(c)

sin260◦−sin230◦= √3

2!2

−1

22

4−1

4=1

sin (60◦+ 30◦) sin (60◦−30◦) = sin 90◦·sin 30◦= 1 ·1

2=1

This identity is correct for the given angles.

4. See Chapter 2, Section 12 and the Appendix of Chapter 2 to review some of

the geometry used in this solution.

(a) Because ABC is subtended by the diameter AC,ABC is a right

angle and △ABC is a right triangle (this fact is known as Thales’s

Theorem). Therefore, sin αis equal to the length of the opposite side

(BC) divided by the length of the hypotenuse (AC). AC is a diameter

of the circle, so it has length 1. Thus, we have that sin αis simply equal

to BC.

A similar argument shows that △ADC is a right triangle with hy-

potenuse AC of length 1, which implies that sin β=DC

(b) Recall that chords of congruent circles which subtend equal angles are

themselves equal. This implies that BC in the diagram of part (a) is

equal to BC in the diagram of part (b) because in both diagrams, the

chord BC subtends an angle of measure α. Similarly, DC is the same

in both diagrams because in both diagrams, the chord DC subtends an

angle of measure β. Therefore, BC is still equal to sin α, and DC is

still equal to sin β.

inscribed angle with measure αin a circle with diameter 1 has length

sin α. Thus, we draw BD, the chord which subtends BAD in both

ﬁgures and which consequently has length sin (α+β).

Note that the above reasoning implies that the sine of an angle with measure

less than 180◦cannot exceed 1 since the diameter is the longest chord in a

circle.

5. Recall that the sine of any angle is at most 1. Therefore,

sin 105◦≤1 = 1

2+1

2<sin 45◦+ sin 60◦,

which shows that sin 105◦cannot equal sin 45◦+ sin 60◦.

Page 125

1. Addition formula for sine:

sin (60◦+ 30◦) = sin 60◦cos 30◦+ cos 60◦sin 30◦

=√3

2·√3

2+1

2·1

4+1

= 1

= sin 90◦

Addition formula for cosine:

cos (60◦+ 30◦) = cos 60◦cos 30◦−sin 60◦sin 30◦

2·√3

2−√3

2·1

=√3

4−√3

= 0

= cos 90◦

Diﬀerence formula for sine:

sin (60◦−30◦) = sin 60◦cos 30◦−cos 60◦sin 30◦

=√3

2·√3

2−1

2·1

4−1

= sin 30◦

Diﬀerence formula for cosine:

cos (60◦−30◦) = cos 60◦cos 30◦+ sin 60◦sin 30◦

2·√3

2+√3

2·1

=√3

4+√3

=√3

= cos 30◦

2. Addition formula for sine (α= 0):

sin (0 + β) = sin 0 cos β+ cos 0 sin β

= 0 ·cos β+ 1 ·sin β

= sin β

Addition formula for cosine (α= 0):

cos (0 + β) = cos 0 cos β−sin 0 sin β

= 1 ·cos β−0·sin β

= cos β

Diﬀerence formula for sine (α= 0):

sin (0 −β) = sin 0 cos β−cos 0 sin β

= 0 ·cos β−1·sin β

=−sin β

Notice that this demonstrates that the sine function is odd.

Diﬀerence formula for cosine (α= 0):

cos (0 −β) = cos 0 cos β+ sin 0 sin β

= 1 ·cos β+ 0 ·sin β

= cos β

Notice that this demonstrates that the cosine function is even.

Addition formula for sine (β= 0):

sin (α+ 0) = sin αcos 0 + cos αsin 0

= sin α·1 + cos α·0

= sin α

Addition formula for cosine (β= 0):

cos (α+ 0) = cos αcos 0 −sin αsin 0

= cos α·1−sin α·0

= cos α

Diﬀerence formula for sine (β= 0):

sin (α−0) = sin αcos 0 −cos αsin 0

= sin α·1−cos α·0

= sin α

Diﬀerence formula for cosine (β= 0):

cos (α+ 0) = cos αcos 0 + sin αsin 0

= cos α·1 + sin α·0

= cos α

3. Following the hint, we notice that in a right triangle, the side opposite one

of the acute angles is the side adjacent to the other acute angle. Thus, if

α+β=π/2, then sin α= cos βand sin β= cos α(see also Chapter 1,

Section 4).

sin (α+β) = sin αcos β+ cos αsin β

= sin αsin α+ cos αcos α

= sin2α+ cos2α

= 1

4. Addition formula for sine:

sin π

4+π

4= sin π

4cos π

4+ cos π

4sin π

=√2

2·√2

2+√2

2·√2

2+1

= 1

= sin π

Addition formula for cosine:

cos π

4+π

4= cos π

4cos π

4−sin π

4sin π

=√2

2·√2

2−√2

2·√2

2−1

= 0

= cos π

5. Recall that (A±B)2=A2±2AB +B2.

(sin αcos β+ cos αsin β)2+ (cos αcos β−sin αsin β)2

= sin2αcos2β+ 2 sin αcos βcos αsin β+ cos2αsin2β+

cos2αcos2β−2 cos αcos βsin αsin β+ sin2αsin2β

= sin2αcos2β+ cos2αsin2β+ cos2αcos2β+ sin2αsin2β

= sin2αcos2β+ sin2β+ cos2αsin2β+ cos2β

= sin2α·1 + cos2α·1

= 1

6. After expanding using the identity (A+B) (A−B) = A2−B2, we cleverly

“add by zero” to get the desired result.

(sin αcos β+ cos αsin β) (sin αcos β−cos αsin β)

= sin2αcos2β−cos2αsin2β

= sin2αcos2β+ sin2αsin2β−sin2αsin2β−cos2αsin2β

= sin2αcos2β+ sin2β−sin2βsin2α+ cos2α

= sin2α·1−sin2β·1

= sin2α−sin2β

Page 129

Page 131

sin (30◦+ 30◦) = sin 30◦cos 30◦+ cos 30◦sin 30◦

2·√3

2+√3

2·1

=√3

4+√3

=√3

= sin 60◦

cos (30◦+ 30◦) = cos 30◦cos 30◦−sin 30◦sin 30◦

=√3

2·√3

2−1

2·1

4−1

= cos 60◦

2. Assuming αand βare acute angles:

sin α=3

5=⇒cos α=p1−sin2α=q1−(3/5)2=4

sin β=5

13 =⇒cos β=q1−sin2β=q1−(5/13)2=12

sin (α+β) = sin αcos β+ cos αsin β

5·12

13 +4

5·5

=36

65 +20

=56

cos (α+β) = cos αcos β−sin αsin β

5·12

13 −3

5·5

=48

65 −15

=33

sin 75◦= sin (45◦+ 30◦)

= sin 45◦cos 30◦+ cos 45◦sin 30◦

=√2

2·√3

2+√2

2·1

=√6 + √2

cos 75◦= cos (45◦+ 30◦)

= cos 45◦cos 30◦−sin 45◦sin 30◦

=√2

2·√3

2−√2

2·1

=√6−√2

sin 15◦= sin (45◦−30◦)

= sin 45◦cos 30◦−cos 45◦sin 30◦

=√2

2·√3

2−√2

2·1

=√6−√2

cos 15◦= cos (45◦−30◦)

= cos 45◦cos 30◦+ sin 45◦sin 30◦

=√2

2·√3

2+√2

2·1

=√6 + √2

Notice that 75◦and 15◦are complementary angles, so we know sin 75◦=

cos 15◦and sin 15◦= cos 75◦.

5. (a) Yes, let α=β=π/4.

cos π

4+π

4= cos π

4cos π

4−sin π

4sin π

=√2

2·√2

2−√2

2·√2

2−1

= 0

More generally, we could suppose α+β=π/2 and follow the approach

in Exercise 3 of Section 2 earlier in this chapter.

(b) If αand βare acute angles, then 0 < α +β < π. Using the unit circle,

we can see that sin (α+β) must be positive since the angle α+βlies

in the upper-half of the plane, where the sine function is positive.

angles since the sum and product of positive real numbers is also posi-

tive.

cos (α+β) need not be positive. As shown in part (a) of this exercise,

cos (α+β) can equal 0. Furthermore, cos (α+β) can be negative. Let

α=β=π/3. Then, assuming that we can extend the cosine addition

formula to angles αand βsuch that α+βis obtuse,

cos π

3+π

3= cos π

3cos π

3−sin π

3sin π

2·1

2−√3

2·√3

4−3

=−1

6. As we saw in Exercise 1 of Section 1 of this chapter, sin α+ sin βdoes

not equal sin (α+β) in general. A similar table can be used to show that

sin α−sin βdoes not equal sin (α−β) in general.

7. This is not a coincidence. The identity holds true even when substituting

more “arbitrary” values in for αand β. For example, using α= 37◦and

β= 19◦, we ﬁnd that both sin2α−sin2βand sin (α+β) sin (α−β) are

equal to approximately 0.2562.

8. You may also refer to the proof in Exercise 6 of Section 2 of this chapter.

sin (α+β) sin (α−β) = (sin αcos β+ cos αsin β) (sin αcos β−cos αsin β)

= sin2αcos2β−cos2αsin2β

= sin2αcos2β+ sin2αsin2β−sin2αsin2β−cos2αsin2β

= sin2αcos2β+ sin2β−sin2βsin2α+ cos2α

= sin2α·1−sin2β·1

= sin2α−sin2β

9. This proof is nearly identical to the one in the previous part. We just make a

small modiﬁcation in how we “add by zero” in order to obtain to the desired

result.

sin (α+β) sin (α−β) = (sin αcos β+ cos αsin β) (sin αcos β−cos αsin β)

= sin2αcos2β−cos2αsin2β

= sin2αcos2β+ cos2αcos2β−cos2αcos2β−cos2αsin2β

= cos2βsin2α+ cos2α−cos2αcos2β+ sin2β

= cos2β·1−cos2α·1

= cos2β−cos2α

10. We apply the sine addition formula in reverse.

sin 18◦cos 12◦+ cos 18◦sin 12◦= sin (18◦+ 12◦)

= sin 30◦

11. (a) Since we have not proved that the sine addition formula works for all

angles αand β, we use properties of the sine and cosine functions to

avoid working with angles larger than 90◦.

sin 113◦cos 307◦+ cos 113◦sin 307◦

= sin (180◦−67◦) cos (360◦−53◦) + cos (180◦−67◦) sin (360◦−53◦)

= sin 67◦cos 53◦+ (−cos 67◦) (−sin 53◦)

= sin (67◦+ 53◦)

= sin 120◦

= sin 60◦

=√3

(b) Plugging into a calculator,

sin 113◦cos 307◦+ cos 113◦sin 307◦≈0.866 ≈√3

2= sin 60◦

we arrive at the same result.

sin 113◦cos 307◦+ cos 113◦sin 307◦= sin (113◦+ 307◦)

= sin 420◦

= sin 60◦

=√3

12. We can use the addition formulas for sine and cosine by rewriting 2αas

α+α.

sin 2αcos α−cos 2αsin α= sin (α+α) cos α−cos (α+α) sin α

= (sin αcos α+ cos αsin α) cos α−(cos αcos α−sin αsin α) sin α

= sin αcos2α+ sin αcos2α−sin αcos2α+ sin3α

= sin αcos2α+ sin3α

= sin αcos2α+ sin2α

= sin α

13.

sin (α+β) sin β+ cos (α+β) cos β= (sin αcos β+ cos αsin β) sin β+ (cos αcos β−sin αsin β) cos β

= sin αsin βcos β+ sin2βcos α+ cos αcos2β−sin αsin βcos β

= sin2βcos α+ cos αcos2β

= cos αsin2β+ cos2β

= cos α

14.

sin (α+β)−cos αsin β

cos (α+β) + sin αsin β=sin αcos β+ cos αsin β−cos αsin β

cos αcos β−sin αsin β+ sin αsin β

=sin αcos β

cos αcos β

=sin α

cos α

= tan α

15.

sin α+π

4= sin αcos π

4+ cos αsin π

= sin α·√2

2+ cos α·√2

=√2

2(sin α+ cos α)

16.

cos (α+β)

cos αcos β=cos αcos β−sin αsin β

cos αcos β

= 1 −sin αsin β

cos αcos β

= 1 −sin α

cos α·sin β

cos β

= 1 −tan αtan β

17. Applying the law of cosines, we have that (b1+b2)2=c2

1+c2

2−2c1c2cos (α+β).

Solving for cos (α+β), we get

cos (α+β) = c2

1+c2

2−(b1+b2)2

2c1c2

Before proceeding further, let’s establish some relationships between the

variables in the diagram. First, by the Pythagorean theorem, we have that

h2=c2

1−b2

1=c2

2−b2

2. Additionally, we can compute the sines and cosines

for the angles αand β:

sin α=b1

,sin β=b2

,cos α=h

,cos β=h

We can now simplify our expression for cos (α+β).

cos (α+β) = c2

1+c2

2−(b1+b2)2

2c1c2

=c2

1+c2

2−b2

1−2b1b2−b2

2c1c2

=2h2−2b1b2

2c1c2

=h2−b1b2

c1c2

c1·h

c2−b1

c1·b2

= cos αcos β−sin αsin β

Chapter 7: Trigonometric Identities

Page 141

1. (a) Yes

(b) No

(d) Yes

(e) No

(f) No

2. (a)

tan α=sin α

cos α

(b)

(1 + tan α) (1 −tan α)=1−tan2α

= 1 −sin2α

cos2α

=cos2α−sin2α

cos2α

(c)

tan α+ tan β

1−tan αtan β=tan α+ tan β

1−tan αtan β·cos αcos β

cos αcos β

=sin αcos β+ cos αsin β

cos αcos β−sin αsin β

(d)

tan2α+ cot2α=sin2α

cos2α+cos2α

sin2α

=sin4α+ cos4α

sin2αcos2α

(e)

tan αcot α=sin α

cos α·cos α

sin α

= 1

(f)

1 + tan2α= 1 + sin2α

cos2α

=cos2α+ sin2α

cos2α

3. The Principle of Analytic Continuation does not apply because p1−sin2α

is not a rational trigonometric function. The identity is incorrect for α=

2π/3 as cos (2π/3) = −1/2, while q1−sin2(2π/3) = 1/2.

4. The Principle of Analytic Continuation does apply because both sin2α+

cos2αand 1 are rational trigonometric functions. The identity is correct for

α= 2π/3.

sin22π

3+ cos22π

3= √3

2!2

+1

22

4+1

4= 1

Page 142

1. Since αand βare acute angles,

sin α=3

5=⇒cos α=p1−sin2α=s1−3

52

sin β=5

13 =⇒cos β=q1−sin2β=s1−5

132

=12

sin (α+β) = sin αcos β+ cos αsin β

5·12

13 +4

5·5

=36

65 +20

=56

cos (α+β) = cos αcos β−sin αsin β

5·12

13 −3

5·5

=48

65 −15

=33

α+βlies in the ﬁrst quadrant because sin (α+β) and cos (α+β) are both

positive.

2. Since αand βare acute angles,

sin α=4

5=⇒cos α=p1−sin2α=s1−4

52

sin β=12

13 =⇒cos β=q1−sin2β=s1−12

132

sin (α+β) = sin αcos β+ cos αsin β

5·5

13 +3

5·12

=20

65 +36

=56

cos (α+β) = cos αcos β−sin αsin β

5·5

13 −4

5·12

=15

65 −48

=−33

α+βlies in the second quadrant because sin (α+β) is positive and cos (α+β)

is negative.

sin α=3

5=⇒cos α=±p1−sin2α=±s1−3

52

=±4

sin β=5

13 =⇒cos β=±q1−sin2β=±s1−5

132

=±12

cos α > 0, cos β > 0:

sin (α+β) = sin αcos β+ cos αsin β

5·12

13 +4

5·5

=36

65 +20

=56

cos α > 0, cos β < 0:

sin (α+β) = sin αcos β+ cos αsin β

5−12

13+4

5·5

=−36

65 +20

=−16

cos α < 0, cos β > 0:

sin (α+β) = sin αcos β+ cos αsin β

5·12

13 +−4

5·5

=36

65 −20

=16

cos α < 0, cos β < 0:

sin (α+β) = sin αcos β+ cos αsin β

5−12

13+−4

5·5

=−36

65 −20

=−56

There are four possible answers for sin (α+β).

4. (a)

sin 2π

3cos π

3−cos 2π

3sin π

3=√3

2·1

2−−1

2√3

=√3

4+√3

=√3

= sin π

3

= sin 2π

3−π

3

(b)

sin π

4cos 3π

4−cos π

4sin 3π

4=√2

2 −√2

2!−√2

2·√2

=−1

2−1

=−1

= sin −π

2

= sin π

4−3π

4

(c)

sin −π

6cos 3π

2−cos −π

6sin 3π

2=−1

2·0−√3

2(−1)

=√3

= sin π

3

= sin −5π

3

= sin −π

6−3π

2

5. Applying the identity

(A−B)2+ (A+B)2=A2−2AB +B2+A2+ 2AB +B2= 2A2+ 2B2,

we have that,

cos2(γ+δ) + cos2(γ−δ) = (cos γcos δ−sin γsin δ)2+ (cos γcos δ+ sin γsin δ)2

= 2 cos2γcos2δ+ 2 sin2γsin2δ.

Therefore,

cos2α+ cos22π

3+α+ cos22π

3−α= cos2α+ 2 cos22π

3cos2α+ 2 sin22π

3sin2α

= cos2α+ 2 1

4cos2α+ 2 3

4sin2α

2cos2α+ sin2α

sin (x+y) + sin (x−y) = sin xcos y+ cos xsin y+ sin xcos y−cos xsin y

= sin xcos y+ sin xcos y

= 2 sin xcos y

cos (x+y) + cos (x−y) = cos xcos y−sin xsin y+ cos xcos y+ sin xsin y

= cos xcos y+ cos xcos y

= 2 cos xcos y

8. Since (A−B)(A+B) = A2−B2,

cos (x+y) cos (x−y) = (cos xcos y−sin xsin y) (cos xcos y+ sin xsin y)

= cos2xcos2y−sin2xsin2y

9. Since (A+B)(A−B) = A2−B2,

sin (x+y) sin (x−y) = (sin xcos y+ cos xsin y) (sin xcos y−cos xsin y)

= sin2xcos2y−cos2xsin2y

10.

cos (x+y) cos (x−y)−sin (x+y) sin (x−y) = cos2xcos2y−sin2xsin2y−sin2xcos2y−cos2xsin2y

= cos2xcos2y+ sin2y−sin2xsin2y+ cos2y

= cos2x−sin2x

11.

cos 2x= cos (x+x)

= cos xcos x−sin xsin x

= cos2x−sin2x

There is no error, because cos 2x= cos2x−sin2x.

12.

cos (α+β) cos β+ sin (α+β) sin β= (cos αcos β−sin αsin β) cos β+ (sin αcos β+ cos αsin β) sin β

= cos αcos2β−sin αsin βcos β+ sin αsin βcos β+ sin2βcos α

= cos αcos2β+ sin2βcos α

= cos αcos2β+ sin2β

= cos α

Alternatively, by applying the cosine diﬀerence formula in reverse,

cos (α+β) cos β+ sin (α+β) sin β= cos (α+β−β) = cos α

Page 144

tan 7π

6+5π

3=

tan 7π

6+ tan 5π

1−tan 7π

6tan 5π

=1/√3−√3

1−1/√3−√3

=1/√3−√3

2·√3

√3

=1−3

2√3

=−1

√3

= tan −π

6

= tan 17π

6

= tan 7π

6+5π

3

2. Because the tangent function is odd, we know tan −β=−tan β.

tan (α−β) = tan (α+ (−β))

=tan α+ tan −β

1−tan αtan −β

=tan α−tan β

1 + tan αtan β

tan π

4+α=

tan π

4+ tan α

1−tan π

4tan α

=1 + tan α

1−tan α

tan π

4−α=

tan π

4−tan α

1 + tan π

4tan α

=1−tan α

1 + tan α

5. Since β=π/4−α,

(1 + tan α) (1 + tan β) = (1 + tan α)1 + tan π

4−α

= (1 + tan α)1 + 1−tan α

1 + tan α

= 1 + tan α+ 1 −tan α

= 2

Alternatively, since tan (α+β) = tan π/4 = 1,

(1 + tan α) (1 + tan β) = 1 + tan α+ tan β+ tan αtan β

= 1 + (tan α+ tan β)1−tan αtan β

1−tan αtan β+ tan αtan β

= 1 + (1 −tan αtan β)tan α+ tan β

1−tan αtan β+ tan αtan β

= 1 + (1 −tan αtan β) tan (α+β) + tan αtan β

= 1 + 1 −tan αtan β+ tan αtan β

= 2

tan (α+β+γ) = tan (α+β) + tan γ

1−tan (α+β) tan γ

tan α+ tan β

1−tan αtan β+ tan γ

1−tan α+ tan β

1−tan αtan βtan γ

tan α+ tan β

1−tan αtan β+ tan γ

1−tan α+ tan β

1−tan αtan βtan γ·1−tan αtan β

1−tan αtan β

=tan α+ tan β+ tan γ(1 −tan αtan β)

1−tan αtan β−(tan α+ tan β) tan γ

=tan α+ tan β+ tan γ−tan αtan βtan γ

1−tan αtan β−tan αtan γ−tan βtan γ

7. Since tan (α+β+γ) = tan π= 0, from the previous part, we have

tan α+tan β+tan γ−tan αtan βtan γ= 0 =⇒tan α+tan β+tan γ= tan αtan βtan γ.

This is because if a fraction equals zero, its numerator must be zero.

8. First, applying the tangent addition formula,

tan 3α= tan (2α+α) = tan 2α+ tan α

1−tan 2αtan α=⇒tan 2α+tan α= tan 3α(1 −tan 2αtan α).

Therefore,

tan 3α−tan 2α−tan α= tan 3α−tan 3α(1 −tan 2αtan α) = tan 3αtan 2αtan α.

tan αis not deﬁned for α=π

2+nπ. Therefore, tan 2αis not deﬁned for

α=π

4+nπ

2and tan 3αis not deﬁned for α=π

6+nπ

Page 147

1. (a) Since sin2α+ cos2α= 1, we know cos α=q1−(7

25 )2(and cannot be

the negative version because cos αis given as positive).

Thus sin 2α= 2 sin αcos α= 2 ·7

25 ·q1−(7

25 )2=336

625 .

And cos 2α= 1 −2 sin2α= 1 −2( 7

25 )2=527

625 .

(b) This part is similar except that we use the negative version of cosine,

namely cos α=−q1−(7

25 )2.

Thus sin 2α= 2 sin αcos α=−336

625 .

However, cosine value remains the same: cos 2α= 1 −2 sin2α= 1 −

2( 7

25 )2=527

625 .

2. Firstly, sin 2α= 2 sin αcos α. In other words, the sine value will be a product

of rational numbers, so it will also be rational. Similarly, cos 2α= 2 cos2α−1

will be rational because cos αis rational. Exercise 1 conﬁrms this result.

3. Let’s use the double angle formula cos 2α= 2 cos2α−1.

cos 2α= cos2α=⇒2 cos2α−1 = cos2α=⇒cos2α= 1 =⇒cos α=±1

cos αhas a magnitude of 1 precisely when αis an integer multiple of π, so

the student’s angle must have also been an integer multiple of π.

4. We start with the given equation:

sin α+ cos α= 0.2

sin2α+ 2 sin αcos α+ cos2α= 0.04 (Squared both sides.)

1 + 2 sin αcos α= 0.04

2 sin αcos α=−0.96

Note that that is simply sin 2α. Hooray!

5. We can follow a very similar strategy here to ﬁnd 1 −2 sin αcos α= 0.09.

Then sin 2α= 2 sin αcos α= 0.91.

cos 2αcos α+ sin 2αsin α=2 cos2α−1cos α+ 2 sin αcos αsin α

= cos α2 cos2α−1 + 2 sin2α

= cos α(2 −1)

= cos α

Alternatively, we may use the cosine diﬀerence formula.

cos 2αcos α+ sin 2αsin α= cos (2α−α) = cos α

7. Applying the sine addition formula,

sin 2αcos α+ cos 2αsin α= sin (2α+α) = sin 3α

Applying the sine subtraction formula,

sin 4αcos α−cos 4αsin α= sin (4α−α) = sin 3α

Since both sides of the identity are equal to sin 3α, the identity is true.

8. Yes, the book asked you to prove something incorrect! For counterexample,

consider that cos 2αcan be negative but cos2αis never negative. However,

we can prove a relationship.

Recall that cos 2α= cos2α−sin2α. However, sin2α≥0 since it’s a square.

sin2α≥0

−sin2α≤0

cos2α−sin2α≤cos2α

cos 2α≤cos2α

sin α

2−cos α

22

= sin2α

2−2 sin α

2cos α

2+ cos2α

= 1 −sin α

10. Following the hint, we compute the value of cos 10◦sin 10◦sin 50◦sin 70◦.

Mcos 10◦= cos 10◦sin 10◦sin 50◦sin 70◦

2sin 20◦sin 50◦sin 70◦

2cos 70◦sin 50◦sin 70◦

4sin 140◦sin 50◦

4sin 40◦sin 50◦

4cos 50◦sin 50◦

8sin 100◦

8sin 80◦

8cos 10◦

Since Mcos 10◦=1

8cos 10◦, we have that the value of the original expression

Mis equal to 1

11. We begin by computing sin 20◦cos 20◦cos 40◦cos 80◦.

sin 20◦cos 20◦cos 40◦cos 80◦=1

2sin 40◦cos 40◦cos 80◦

4sin 80◦cos 80◦

8sin 160◦

8sin 20◦

This implies that cos 20◦cos 40◦cos 80◦= 1/8.

12. We begin by computing cos π/10 sin π/10 sin π/5.

cos π

10 sin π

10 cos π

5=1

2sin π

5cos π

4sin 2π

4cos π

This implies that sin π/10 cos π/5=1/4.

Page 148

cos 3α= cos (2α+α)

= cos 2αcos α−sin 2αsin α

=2 cos2α−1cos α−2 sin αcos αsin α

= 2 cos3α−cos α−2 sin2αcos α

= 2 cos3α−cos α−21−cos2αcos α

= 4 cos3α−3 cos α

sin 3α= 3 sin α−4 sin3α

= 3 3

5−43

53

5−4·27

125

=117

125

If sin α= 3/5, then cos α=±4/5. Therefore,

cos 3α= 4 cos3α−3 cos α

= 4 ±4

53

−3±4

5

=±256

125 ∓12

=∓44

125

3. If cos α= 4/5, then sin α=±3/5. Therefore,

sin 3α= 3 sin α−4 sin3α

= 3 ±3

5−4±3

53

=±9

5∓4·27

125

=±117

125

cos 3α= 4 cos3α−3 cos α

= 4 4

53

−34

5

=256

125 −12

=−44

125

4. (a)

cos 4α= 2 cos22α−1

= 2 2 cos2α−12−1

= 2 4 cos4α−4 cos2α+ 1−1

= 8 cos4α−8 cos2α+ 1

(b)

cos 4α= 1 −2 sin22α

= 1 −8 sin2αcos2α

= 1 −8 sin2α1−sin2α

= 1 −8 sin2α+ 8 sin4α

sin 3αcos α−cos 3αsin α=3 sin α−4 sin3αcos α−4 cos3α−3 cos αsin α

= 3 sin αcos α−4 sin3αcos α−4 cos3αsin α+ 3 sin αcos α

= 2 sin αcos α3−2 sin2α−2 cos2α

= 2 sin αcos α(3 −2)

= sin 2α

Alternatively, applying the sine diﬀerence formula,

sin 3αcos α−cos 3αsin α= sin (3α−α) = sin 2α

sin 3α

sin α−cos 3α

cos α=3 sin α−4 sin3α

sin α−4 cos3α−3 cos α

cos α

= 3 −4 sin2α−4 cos2α−3

= 6 −4 sin2α−4 cos2α

= 2

Alternatively, using the result from the previous exercise,

sin 3α

sin α−cos 3α

cos α=sin 3αcos α−cos 3αsin α

sin αcos α

=sin 2α

sin αcos α

=2 sin αcos α

sin αcos α

= 2

7. (a) Applying the result of Exercise 9 from Section 3 of this chapter,

4 sin αsin (60◦+α) sin (60◦−α) = 4 sin αsin260◦cos2α−cos260◦sin2α

= 4 sin α3

4cos2α−1

4sin2α

= 3 sin αcos2α−sin3α

= 3 sin α1−sin2α−sin3α

= 3 sin α−4 sin3α

= sin 3α

(b) Applying the result of Exercise 8 from Section 3 of this chapter,

4 cos αcos (60◦+α) cos (60◦−α) = 4 cos αcos260◦cos2α−sin260◦sin2α

= 4 cos α1

4cos2α−3

4sin2α

= cos3α−3 sin2αcos α

= cos3α−31−cos2αcos α

= 4 cos3α−3 cos α

sin 4α= 2 sin 2αcos 2α

= 4 sin αcos α2 cos2α−1

= 8 sin αcos3α−4 sin αcos α

=⇒sin 4α

sin α= 8 cos3α−4 cos α

9. In the penultimate step, we apply the following identity:

(A−B)3=A3−3A2B+ 3AB2−B3,

taking A= cos2αand B= sin2α.

sin 3αsin3α+ cos 3αcos3α=3 sin α−4 sin3αsin3α+4 cos3α−3 cos αcos3α

= 3 sin4α−4 sin6α+ 4 cos6α−3 cos4α

= 3 sin4α−4 sin4α1−cos2α+ 4 cos4α1−sin2α−3 cos4α

= sin4α4 cos2α−1+ cos4α1−4 sin2α

= sin4α4 cos2α−sin2α−cos2α+ cos4αsin2α+ cos2α−4 sin2α

= 3 sin4αcos2α−sin6α+ cos6α−3 cos4αsin2α

=cos2α−sin2α3

= cos32α

Page 150

cos α= 1 =⇒cos α

2=±r1+1

2=±1

cos αis equal to 1 when αis an integer multiple of 2π. If α= 4nπ for some

integer n(i.e., an even integer multiple of 2π), then cos α/2 = cos 2nπ = 1.

Otherwise, if α= (4n+ 2) πfor some integer n(i.e., an odd integer multiple

of 2π), then cos α/2 = cos (2n+ 1) π=−1.

For a particular example of each case, we may set n= 0.

α= 4 (0) π= 0:

cos 0 = 1,cos 0

2= cos 0 = 1

α= (4 (0) + 2) π= 2π:

cos 2π= 1,cos 2π

2= cos π=−1

2. (a) We take the positive square root because 60◦/2 = 30◦lies in the ﬁrst

quadrant.

cos 60◦

2=r1 + cos 60◦

=r1+1/2

=r3

=√3

(b) We take the positive square root because 120◦/2 = 60◦lies in the ﬁrst

quadrant.

cos 120◦

2=r1 + cos 120◦

=r1−1/2

=r1

second quadrant.

cos 240◦

2=−r1 + cos 240◦

=−r1−1/2

=−r1

=−1

αQuadrant α?α/2 Quadrant α/2? cos α/2

780◦I 390◦I√3/2

1020◦IV 510◦II −√3/2

1140◦I 570◦III −√3/2

1380◦IV 690◦IV √3/2

−60◦IV −30◦IV √3/2

−300◦I−150◦III −√3/2

−420◦IV −210◦II −√3/2

−660◦I−330◦I√3/2

−780◦IV −390◦IV √3/2

sin 15◦=r1−cos 30◦

=s1−√3/2

=s2−√3

=p2−√3

cos 15◦=r1 + cos 30◦

=s1 + √3/2

=s2 + √3

=p2 + √3

5. Because |cos α| ≤ 1, 1 ±cos α≥0, so the expressions under the square roots

in the sine and cosine half-angle formulas will not be negative.

6. The square root sign in the half-angle formula prevents it from being a

rational trigonometric function, so the Principle of Analytic Continuation

does not apply.

7. (a)

tan α

2tan β

2+ tan α

2tan γ

2+ tan β

2tan γ

2= tan α

2tan β

2+ tan γ

2tan α

2+ tan β

2

= tan α

2tan β

2+ tan γ

2tan α+β

21−tan α

2tan β

2

= tan α

2tan β

2+ 1 −tan α

2tan β

= 1

Alternatively, we can recall the extended tangent addition formula de-

rived in Exercise 6 of Section 4 of this chapter:

tan (α+β+γ) = tan α+ tan β+ tan γ−tan αtan βtan γ

1−tan αtan β−tan αtan γ−tan βtan γ

Since α/2 + β/2 + γ/2 = π/2, tan (α/2 + β/2 + γ/2) is undeﬁned. This

implies that the denominator of the tangent addition formula is 0 (as-

suming that none of tan α/2, tan β/2, or tan γ/2 are undeﬁned). There-

fore, we can conclude

tan α

2tan β

2+ tan α

2tan γ

2+ tan β

2tan γ

2= 1.

(b) Following a similar approach as to the previous part, we begin by noting

that sin α+β

2= cos γ

4 cos α

2cos β

2cos γ

2= 4 cos α

2cos β

2sin α+β

= 4 cos α

2cos β

2sin α

2cos β

2+ cos α

2sin β

2

= 4 sin α

2cos α

2cos2β

2+ 4 sin β

2cos β

2cos2α

= 2 sin αcos2β

2+ 2 sin βcos2α

= 2 sin α1 + cos β

2+ 2 sin β1 + cos α

2

= sin α+ sin αcos β+ sin β+ sin βcos α

= sin α+ sin β+ sin (α+β)

= sin α+ sin β+ sin (π−α−β)

= sin α+ sin β+ sin γ

Page 152

1. Because 1 + cos αis non-negative, division by 1 + cos αdoes not change the

sign of sin α, which means tan (α/2) and sin α/ (1 + cos α) have the same

sign.

tan α

2=±r1−cos α

1 + cos α

=±r1−cos α

1 + cos α·1−cos α

1−cos α

=±s(1 −cos α)2

1−cos2α

=±s(1 −cos α)2

sin2α

=±1−cos α

sin α

For acute angles α, we need to take the positive branch of the square root so

that the signs of both sides of the half-angle formula agree. By the Principle

of Analytic Continuation, since the positive branch is correct for all acute

angles and both sides of the formula are rational trigonometric expressions,

it is correct for all angles in general, so we have

tan α

2=1−cos α

sin α

Page 153

2cos (α−β)−1

2cos (α+β) = 1

2(cos αcos β+ sin αsin β−cos αcos β+ sin αsin β)

2(2 sin αsin β)

= sin αsin β

2sin (α+β) + 1

2sin (α−β) = 1

2(sin αcos β+ cos αsin β+ sin αcos β−cos αsin β)

2(2 sin αcos β)

= sin αcos β

sin 75◦sin 15◦=1

2cos (75◦−15◦)−1

2cos (75◦+ 15◦)

2cos 60◦−1

2cos 90◦

2·1

2−1

2·0

sin 75◦cos 15◦=1

2sin (75◦+ 15◦) + 1

2sin (75◦−15◦)

2sin 90◦+1

2sin 60◦

2·1 + 1

2·√3

=2 + √3

4. (a)

cos 75◦cos 15◦=1

2cos (75◦+ 15◦) + 1

2cos (75◦−15◦)

2cos 90◦+1

2cos 60◦

2·0 + 1

2·1

Alternatively, by the cosine subtraction formula, cos 60◦= cos 75◦cos 15◦+

sin 75◦sin 15◦so

cos 75◦cos 15◦= cos 60◦−sin 75◦sin 15◦=1

2−1

4=1

Alternatively, using the idea of sine and cosine being cofunctions,

cos 75◦cos 15◦= sin (90◦−75◦) sin (90◦−15◦) = sin 15◦sin 75◦=1

For the above two alternative solutions, we apply the result from Ex-

ercise 2 above that sin 75◦sin 15◦= 1/4.

(b)

cos 75◦sin 15◦=1

2sin (15◦+ 75◦) + 1

2sin (15◦−75◦)

2sin 90◦−1

2sin 60◦

2·1−1

2·√3

=2−√3

Alternatively, by the sine addition formula, sin 90◦= sin 75◦cos 15◦+

cos 75◦sin 15◦so

cos 75◦sin 15◦= sin 90◦−sin 75◦cos 15◦= 1 −2 + √3

4=2−√3

For the above alternative solution, we apply the result from Exercise 3

above that sin 75◦cos 15◦=2 + √3/4.

2 cos π

4+αcos π

4−α= cos π

4+α+π

4−α+ cos π

4+α−π

4+α

= cos π

2cos 2α

= cos 2α

Alternatively, applying the result of Exercise 8 in Section 3 of this chapter,

2 cos π

4+αcos π

4−α= 2 cos2π

4cos2α−sin2π

4sin2α

= 2 1

2cos2α−1

2sin2α

= cos2α−sin2α

= cos 2α

sin (α+β) sin (α−β) + sin (β+γ) sin (β−γ) + sin (γ+α) sin (γ−α)

2cos 2β−1

2cos 2α+1

2cos 2γ−1

2cos 2β+1

2cos 2α−1

2cos 2γ

= 0

100

sin αsin (β−γ) + sin βsin (γ−α) + sin γsin (α−β)

2cos (α−β+γ)−1

2cos (α+β−γ) + 1

2cos (β−γ+α)−1

2cos (β+γ−α)

2cos (γ−α+β)−1

2cos (γ+α−β)

= 0

Page 155

1. Recall from the previous section that cos (γ+δ) + cos (γ−δ) = 2 cos γcos δ.

Following Example 54, we let γ= (α+β)/2 and δ= (α−β)/2. Substitut-

ing for γand δ, we arrive at the ﬁrst formula,

cos α+ cos β= 2 cos α+β

2cos α−β

Similarly, we recall that cos (γ−δ)−cos (γ+δ) = 2 sin γsin δ. Performing

the same substitution as above, we obtain

cos β−cos α= 2 sin α+β

2sin α−β

2⇐⇒ cos α−cos β=−2 sin α+β

2sin α−β

cos 70◦+ sin 40◦= sin 20◦+ sin 40◦

= 2 sin 20◦+ 40◦

2cos 20◦−40◦

= 2 sin 30◦cos (−10◦)

= 2 ·1

2cos 10◦

= cos 10◦

cos 55◦+ cos 65◦= 2 cos 55◦+ 65◦

2cos 55◦−65◦

= 2 cos 60◦cos (−5◦)

= 2 ·1

2cos 5◦

= cos 5◦

101

cos 20◦+ cos 100◦+ cos 140◦= 2 cos 100◦+ 20◦

2cos 100◦−20◦

2+ cos 140◦

= 2 cos 60◦cos 40◦+ cos 140◦

= 2 ·1

2cos 40◦+ cos 140◦

= cos 40◦+ cos 140◦

= 2 cos 140◦+ 40◦

2cos 140◦−40◦

= 2 cos 90◦cos 50◦

= 0

sin 78◦+ cos 132◦= sin 78◦−cos 48◦

= sin 78◦−sin 42◦

= 2 cos 78◦+ 42◦

2sin 78◦−42◦

= 2 cos 60◦sin 18◦

= 2 ·1

2sin 18◦

= sin 18◦

cos 15◦+ sin 15◦

cos 15◦−sin 15◦=sin 75◦+ sin 15◦

sin 75◦−sin 15◦

=2 sin 45◦cos 30◦

2 cos 45◦sin 30◦

= tan 45◦cot 30◦

=√3

7. (a)

sin (α+β) = sin (π−α−β) = sin γ

(b)

cos (α+β) = −cos (π−α−β) = −cos γ

102

(c)

sin 2α+ sin 2β+ sin 2γ= 2 sin (α+β) cos (α−β) + sin 2γ

= 2 sin γcos (α−β) + 2 sin γcos γ

= 2 sin γ(cos (α−β) + cos γ)

= 2 sin γ(cos (α−β)−cos (α+β))

= 2 sin γ(2 sin αsin β)

= 4 sin αsin βsin γ

sin α+ sin α+2π

3+ sin α+4π

3= 2 sin α+π

3cos −π

3+ sin α+4π

3

= sin α+π

3+ sin α+4π

3

= 2 sin α+5π

6cos −π

2

= 0

9. We ﬁrst note that

sin kα+sin (k+ 2)α= 2 sin (k+ 2)α+kα

2cos (k+ 2)α−kα

2= 2 sin (k+ 1)αcos α.

Therefore,

sin α+ 2 sin 3α+ sin 5α= sin α+ sin 3α+ sin 3α+ sin 5α

= 2 sin 2αcos α+ 2 sin 4αcos α

= 2 cos α(sin 2α+ sin 4α)

= 2 cos α(2 sin 3αcos α)

= 4 cos2αsin 3α.

10.

sin (β−γ)

sin βsin γ+sin (γ−α)

sin γsin α+sin (α−β)

sin αsin β

=sin βcos γ

sin βsin γ−cos βsin γ

sin βsin γ+sin γcos α

sin γsin α−cos γsin α

sin γsin α+sin αcos β

sin αsin β−cos αcos β

sin αsin β

= cot γ−cot β+ cot α−cot γ+ cot β−cot α

= 0

This result also follows from the identity which was proven in Exercise 7 of

the previous section. This can be seen by rewriting the left-hand side of the

equation as a single fraction with a common denominator and noting that

the numerator of this fraction is zero.

103

11.

sin (α−β) + sin (α−γ) + sin (β−γ) = 2 sin 2α−β−γ

2cos γ−β

2+ sin (β−γ)

= 2 sin 2α−β−γ

2cos β−γ

2+ 2 sin β−γ

2cos β−γ

= 2 cos β−γ

2sin 2α−β−γ

2+ sin β−γ

2

= 2 cos β−γ

22 sin α−γ

2cos α−β

2

= 4 cos α−β

2sin α−γ

2cos β−γ

12.

sin (α+β+γ) + sin (α−β−γ) + sin (α+β−γ) + sin (α−β+γ)

= 2 sin αcos (β+γ) + sin (α+β−γ) + sin (α−β+γ)

= 2 sin αcos (β+γ) + 2 sin αcos (β−γ)

= 2 sin α(cos (β+γ) + cos (β−γ))

= 2 sin α(2 cos βcos γ)

= 4 sin αcos βcos γ

Page 158

1. Let β= 2γ.

sin2β+ cos2β= sin22γ+ cos22γ

=2a

1 + a22

+1−a2

1 + a22

=4a2

1+2a2+a4+1−2a2+a4

1+2a2+a4

=1+2a2+a4

1+2a2+a4

= 1

tan 2β=2a

1−a2

1 + a2

1−a2

1 + a2

=sin 2β

cos 2β

104

Page 160

sin α=2·2·3

22+ 32=12

cos α=22−32

22+ 32=−5

These values give the Pythagorean triple 5, 12, 13, provided we take the

absolute value of cos α. However, because cos αis negative, αis not an

acute angle, so it cannot correspond to an angle in a right triangle.

sin α=2·8·5

82+ 52=80

cos α=82−52

82+ 52=39

This corresponds to the right triangle with legs of 39 and 80 and a hypotenuse

of 89.

(2pq)2+q2−p22= 4p2q2+q4−2q2p2+p4=q4+ 2q2p2+p4=q2+p22

The above shows that q2+p2is the hypotenuse.

Page 161 (First)

1. •sin 20◦cos 20◦≈0.3214

•sin 10◦cos 10◦≈0.1710

•sin 5◦cos 5◦≈0.0868

•sin 1◦cos 1◦≈0.0174

•sin 70◦cos 70◦≈0.3214

•sin 80◦cos 80◦≈0.1710

•sin 85◦cos 85◦≈0.0868

•sin 89◦cos 89◦≈0.0174

sin 30◦cos 30◦=1

2·√3

2=√3

sin 45◦cos 45◦=√2

2·√2

2=1

sin 60◦cos 60◦=√3

2·1

2=√3

105

Page 161 (Second)

1. In the below answers for this question, nrepresents an arbitrary integer.

(a)

sin xcos x=1

2=⇒1

2sin 2x=1

=⇒sin 2x= 1

=⇒2x=π

2+ 2nπ

=⇒x=π

4+nπ

(b)

sin xcos x=√3

2=⇒1

2sin 2x=√3

2=⇒sin 2x=√3

Because √3>1, there are no values of xwhich satisfy the given equa-

tion.

(c)

sin xcos x=√3

4=⇒1

2sin 2x=√3

=⇒sin 2x=√3

=⇒2x=π

2±π

6+ 2nπ

=⇒x=π

4±π

12+nπ

2. (c) has no solution because sin xcos x≤0.5<0.6.

3. sin xcos x=Nhas a solution when |N| ≤ 1/2.

sin xcos x=N=⇒1

2sin 2x=N=⇒sin 2x= 2N

2x= arcsin 2Ngives one solution to the above equation. Another (not

necessarily distinct) solution is 2x=π−arcsin 2N. All solutions to the

equation can be generated by adding an integer multiple of 2πto either of

the above angles. Therefore, the general solution to sin xcos x=Nis

x=1

2arcsin 2N+nπ or x=1

2(π−arcsin 2N) + nπ.

Using the properties of cofunctions, this solution can be written in a slightly

more compact manner as follows:

x=π

4±1

2arccos 2N+nπ

106

Page 162 (First)

sin 30◦+ cos 30◦=1

2+√3

2>1

2+1

2= 1

sin 0 + cos 0 = 1

sin π

4+ cos π

4=√2

2+√2

2=√2

Page 162 (Second)

1. Yes, because 1.414 <√2. (Note: sin x+ cos xmust attain the value 1.414

for some xbecause of the Intermediate Value Theorem).

2. No, because 1.415 >√2.

3. We square both sides of the equation to solve for x.

sin x+cos x=√2 =⇒1+sin 2x= 2 =⇒sin 2x= 1 =⇒2x=π

2+2nπ =⇒x=π

4+nπ

4. Since we know the maximum value of (sin x+ cos x)2is 2, the minimum

value of sin x+cos xcannot be less than −√2. The value of −√2 is attained

when x=5π

sin 5π

4+ cos 5π

4=−√2

2−√2

2=−√2

Page 163 (First)

1. sin x+ cos xis maximized when x+π

4=π

2+ 2nπ, where nis an integer.

Therefore,

x=π

4+ 2nπ

2. The minimum value of sin x+ cos xis −√2. This is achieved when

x=−3π

4+ 2nπ.

Page 163 (Second)

1. Yes. Because sin αand cos αare both positive, αmust be in the ﬁrst quad-

rant.

2. The minimum value of 3 sin x+ 4 cos xis −5. This occurs when

x=−α−π

2+ 2nπ.

107

3. Let αbe the positive acute angle such that sin α= 7/√53.

2 sin x+ 7 cos x=√53 2

√53 sin x+7

√53 cos x=√53 sin (α+x)

The above shows that the maximum and minimum values of 2 sin x+ 7 cos x

are √53 and −√53, respectively.

Page 164

cos π

16 =v

t1 + cos π

t1 + 1

2p2 + √2

=s2 + p2 + √2

2r2 + q2 + √2

sin π

16 =v

t1−cos π

t1−1

2p2 + √2

=s2−p2 + √2

2r2−q2 + √2

108

αcos αsin α

2q2 + p2 + √21

2q2−p2 + √2

2r2 + q2 + p2 + √21

2r2−q2 + p2 + √2

2s2 + r2 + q2 + p2 + √21

2s2−r2 + q2 + p2 + √2

128

t2 + s2 + r2 + q2 + p2 + √21

t2−s2 + r2 + q2 + p2 + √2

Page 166

1. 1

2√2≈0.7071

2q2 + √2≈0.9239

2r2 + q2 + √2≈0.9809

2s2 + r2 + q2 + √2≈0.9952

22q2−√2≈3.0615

23r2−q2 + √2≈3.1214

24s2−r2 + q2 + √2≈3.1365

25v

t2−s2 + r2 + q2 + √2≈3.1403

109

3. (a)

cos π

12 =r1 + cos (π/6)

=s1 + √3/2

2q2 + √3

≈0.9659

(b)

cos π

24 =r1 + cos (π/12)

=s1 + p2 + √3/2

2r2 + q2 + √3

≈0.9914

(c)

cos π

48 =r1 + cos (π/24)

t1 + q2 + p2 + √3/2

2s2 + r2 + q2 + √3

≈0.9979

(d)

cos π

96 =r1 + cos (π/48)

t1 + r2 + q2 + p2 + √3/2

t2 + s2 + r2 + q2 + √3

≈0.9995

110

These values approach 1 because for small angles x, cos xgets closer to

1. More formally, the limit of cos xas xapproaches zero is 1.

4. (a)

12 sin π

12 = 6q2−√3≈3.1058

(b)

24 sin π

24 = 12r2−q2 + √3≈3.1326

(c)

48 sin π

48 = 24s2−r2 + q2 + √3≈3.1394

(d)

96 sin π

96 = 48v

t2−s2 + r2 + q2 + √3≈3.1410

Page 168

√1 + √2+1

√2 + √3+. . . +1

√99 + √100 =

k=1

√k+√k+ 1

k=1

√k+√k+ 1 ·√k+ 1 −√k

√k+ 1 −√k

k=1

√k+ 1 −√k

=√2−√1+√3−√2+. . . +√100 −√99

=√100 −√1

= 10 −1

= 9

111

1 + 3 + . . . + (2n+ 1) =

k=0

2k+ 1

k=0

(k+ 1)2−k2

=12−02+22−12+. . . +(n+ 1)2−n2

= (n+ 1)2−02

= (n+ 1)2

=n2+ 2n+ 1

(1 −x)P= (1 −x) (1 + x)1 + x21 + x41 + x81 + x16

=1−x21 + x21 + x41 + x81 + x16 

=1−x41 + x41 + x81 + x16

=1−x81 + x81 + x16

=1−x161 + x16

= 1 −x32

Thus,

P=1−x32

1−x= 1 + x+. . . +x31

sin 20◦P= sin 20◦cos 20◦cos 40◦cos 80◦

2sin 40◦cos 40◦cos 80◦

4sin 80◦cos 80◦

8sin 160◦

8sin 20◦

Thus, P=1

112

Page 170

2 sin x(sin x+ sin 3x+. . . + sin 99x) = 2 sin x

k=0

sin (2k+ 1) x

k=0

2 sin xsin (2k+ 1) x

k=0

cos 2kx −cos (2k+ 2)x

= (cos 0 −cos 2x) + (cos 2x−cos 4x) + . . . + (cos 98x−cos 100x)

= 1 −cos 100x

Thus, the original sum, sin x+. . .+sin 99x, is equal to (1 −cos 100x)/(2 sin x).

Notice that this is equivalent to sin250x/ sin x, which is what would be found

using the formula at the bottom of page 169.

2 sin π

8sin x+. . . + sin x+99π

4= 2 sin π

k=0

sin x+kπ

4

k=0

2 sin π

8sin x+kπ

4

k=0

cos x+k−1

2π

4−cos x+k+1

2π

4

= cos x−π

8−cos x+199π

8

= cos x−π

8−cos x+7π

8

= cos x−π

8+ cos π−x−7π

8

= 2 cos x−π

8

= 2 cos xcos π

8+ 2 sin xsin π

Dividing by 2 sin π

8, we ﬁnd that the original sum is equal to cot π

8cos x+

sin x. Notice that this is equivalent to sin (x+ 99π/8) /sin (π/8), which is

what would be found using the formula at the bottom of page 169.

An alternative way to evaluate this sum is by noticing that the terms of this

sum repeat in periods of 8 because sin x+8π

4= sin (x). Furthermore, the

113

ﬁrst 8 terms of the sum total to zero (try verifying this yourself using the

sine addition formulas). Therefore, to evaluate the whole sum, all we need

to evaluate is the sum of the last four terms:

sin x+96π

4+ sin x+97π

4+ sin x+98π

4+ sin x+99π

4.

By the periodicity in the sum, we can equivalently evaluate the ﬁrst four

terms of the sum. Using the sine addition formulas, we ﬁnd that the sum is

equal to

sin x+ sin x+π

4+ sin x+π

2+ sin x+3π

4

= sin x+√2

2sin x+√2

2cos x+ cos x−√2

2sin x+√2

2cos x

= sin x+ cos x+√2 cos x.

Combining this result with our initial evaluation of this sum shows that

cot π

8= 1 + √2.

3. We begin by multiplying the sum by 2 sin x.

2 sin x(cos 2x+ cos 4x+. . . + cos 2nx) =

k=1

2 sin xcos 2kx

k=1

sin (2k+ 1)x−sin (2k−1)x

= (sin 3x−sin x) + . . . + (sin (2n+ 1)x−sin (2n+ 1)x)

= sin (2n+ 1)x−sin x

Dividing by 2 sin xgives us the value of the sum as

sin(2n+ 1)x

2 sin x−1

4. We begin by multiplying the sum by 2 sin π

2k.

2 sin π

2kcos π

k+ cos 2π

k+. . . + cos nπ

k=

j=1

2 sin π

2kcos jπ

j=1

sin π

kj+1

2−sin π

kj−1

2

= sin nπ

k+π

2k−sin π

= sin nπ

kcos π

2k+ cos nπ

ksin π

2k−sin π

Dividing by 2 sin π

2k, we ﬁnd that the value of the sum is

2sin nπ

kcot π

2k+ cos nπ

k−1.

114

5. The heights of the perpendiculars are given by sin kπ

12 , where kranges from

1 to 11. We consider this sum multiplied by 2 sin π

24 .

k=1

2 sin π

24 sin kπ

12 =

k=1

cos kπ

12 −π

24−cos kπ

12 +π

24

= cos π

24 −cos 23π

= 2 cos π

Dividing this result by 2 sin π

24 gives the value of the sum as cot π

24 .

Alternatively, we can use the formula for series of sines with angles in arith-

metic progression, setting x= 0, n= 11, α=π/12 to ﬁnd that the sum of

the altitudes is

sin 6π

12 sin 11π

sin π

sin 11π

sin π

cos π

sin π

= cot π

24.

Using the half-angle formulas for sine and cosine, we can show that cos π/12 =

2p2 + √3 and sin π/12 = 1

2p2−√3. Therefore,

cot π

24 =

1 + cos π

sin π

1 + 1

2p2 + √3

2p2−√3

=2 + p2 + √3

p2−√3

p2−√3+s2 + √3

2−√3

=2p2−√3

2−√3+2+√3

= 2q2−√32 + √3+2+√3

= 2q2 + √3 + 2 + √3

= 2 r1

2+r3

2!+2+√3

=√2 + √6 + 2 + √3.

115

Let’s elaborate on how to evaluate p2 + √3. Suppose p2 + √3 = √a+

√bfor some rational numbers aand b. Then, squaring both sides, we get

2 + √3 = a+ 2√ab +b. Matching the rational and irrational parts, we get

two equations relating aand b:a+b= 2 and 4ab = 3. Solving this system

shows that aand bare 1/2 and 3/2 (the order doesn’t matter because the

equations are symmetric). Thus, we have shown that

q2 + √3 = r1

2+r3

Chapter 8: Graphs of Trigonometric Functions

Page 177

1. Since k= 5, Period is 5, and Frequency is π

2. Since k=1

4, Period is 1

4, and Frequency is π

1/4= 4π

3. Since k=4

5, Period is 4

5, and Frequency is π

4/5=5π

4. Since k=5

4, Period is 5

4, and Frequency is π

5/4=4π

5. Period is 2π

2π3π

22π

−1

6. Period is 2π

1/3= 6π

1π2π3π4π5π6π

−1

7. Period is 2π

3/2=4π

116

1π2π

−1

8. Period is 2π

2/3= 3π

1π2π3π

−1

9. Period is 2π

2/3= 3π

1π2π3π4π5π6π

−1

10. Period is 2π

3/2=4π

1π2π

−1

11. For y=f(3x)

117

−1 0 1 2 3 4 5 6

−1

For y=f(x

−1 0 1 2 3 4 5 6

−1

Page 179

1. y= 2 sin (x)

1π2π

−2

−1

2. y=1

2sin (x)

1π2π

−1

3. y= 3 sin (2x)

118

1π2π

−3

−2

−1

4. y=1

2sin (3x)

1π2π

−1

5. y= 4 cos (x)

1π2π

−4

−3

−2

−1

6. For y= 3f(x)

−1 0 1 2 3 4 5 6

−3

−2

−1

7. For y=1

3f(x)

119

−1 0 1 2 3 4 5 6

−1

Page 181

1. y= sin x−π

6

1π2π

−1

2. y= sin x+π

6

1π2π

−1

3. y= 2 sin x−π

2

1π2π

−2

−1

4. y=1

2sin x+π

2

120

1π2π

−1

5. y= cos x−π

4

1π2π

−1

6. y= 3 cos x+π

3

1π2π

−3

−2

−1

7. y= sin (x−2x)

1π2π

−1

8. Since the sine wave has shifted right by π

3. The equation is

y= sin x−π

3

121

9. Since the sine wave has shifted right by 2π

3. The equation is

y= sin x−2π

3

10. Since the sine wave has shifted left by π

6. The equation is

y= sin x−π

6

11. Note: this looks like the sine wave has shifted right by π

4, but at that point,

the sine wave is on its wave down.

The actual shift is the xintercept on the far left, calculate its location by

4−π=−3π

4as it’s half a period away, that is, if 2πradians is one period,

then πradians is a half period

Therefore, the sine wave has shifted left by 3π

4. The equation is

y= sin x+3π

4

Page 183

1. y= sin 1

2x−π

6

1π2π3π4π

−1

2. y= sin 1

3x−π

6

1π2π3π4π

−1

3. y= cos 2 x+π

3

122

1π2π3π4π

−1

4. Since period is π, making k=2π

π= 2, making the equation

y= sin 2(x−β)

To solve for β, substitute x= 0, y =−1, as the sine wave passes point

(0,−1), giving

−1 = sin 2(0 −β)

−1 = sin −2β

−π

2=−2β

4=β

β=π

Therefore the equation is

y= sin 2 x−π

4

5. Since half period is 2π, its full period is 4πmaking k=2π

4π=1

2, making the

equation

y= sin 1

2(x−β)

To solve for β, substitute x=5π

6, y = 1, as the sine wave passes point (5π

6,1),

giving

1 = sin 1

25π

6−β

2=1

25π

6−β

π=5π

6−β

6=−β

β=−π

Therefore the equation is

y= sin 1

2x−π

6

123

Page 186

1. Answer A) equal to sin xsince

sin(x+ 2π) = sin(x+ 2π−2π)

= sin(x)

2. Answer C) equal to −sin xsince

sin(x+ 3π) = sin(x+ 3π−2π)

= sin(x+π)

=−sin(x)

3. Answer B) equal to cos xsince

sin x+9

2π= sin x+9

2π−4π

= sin x+π

2

= cos(x)

4. Answer D) equal to −cos xsince

sin x−π

2=−cos(x)

5. Answer B) equal to cos xsince

sin x−3

2π= sin −3

2π+ 2π

= sin x+π

2

= cos(x)

6. Answer D) equal to −cos xsince

sin x+19

2π= sin x+19

2π−10π

= sin x−π

2

=−cos(x)

124

7. Answer D) equal to −cos xsince

−sin x−19

2π=−sin x+19

2π+ 10π

=−sin x+π

2

=−cos(x)

8. Answer B) equal to cos xsince

sin x+157

2π= sin x+157

2π−78π

= sin x+π

2

= cos(x)

9. Answer D) equal to −cos xsince

sin x−157

2π= sin x−157

2π+ 78π

= sin x−π

2

=−cos(x)

10. Note: there is no question 10 in textbook (printing error?)

11. Using deﬁnition that pis called a half-period of function fis f(x+p) =

−f(x), for all values of x which f(x) and f(x+p) are deﬁned

Since cos(x) has period of 2πradians, therefore its half-period is πradians,

or we can compare diagrams

1π2π3π4π

−1

1y= cos(x)

y= cos(x+π)

Which can be seen that cos(x+π) is the same as −cos(x), therefore by

deﬁnition, cos xhas a half-period of πradians

125

For y= tan x

tan(x+π) = sin(x+π)

cos(x+π)

=−sin x

−cos x

= tan x

So πis period of y= tan x

For y= cot x

cot(x+π) = cos(x+π)

sin(x+π)

=−cos x

−sin x

= cot x

So πis period of y= cot x

12. Using deﬁnition that qis called a half-period of function fis f(x+q) =

−f(x), for all values of x which f(x) and f(x+q) are deﬁned

f(x+ 2q) = f(x+q+q)

=−f(x+q)

=−(−f(x))

=f(x)

Therefore if qis half period of some function f, then 2qis a period of f

13. (a)

cos (x+kπ/2) = cos x+π(4n+ 1)

2

= cos (x+ 2nπ +π/2)

= cos(x+π/2)

=−sin x

(b)

cos (x+kπ/2) = cos x+π(4n+ 2)

2

= cos (x+ 2nπ +π)

= cos(x+π)

=−cos x

126

(c)

cos (x+kπ/2) = cos x+π(4n+ 3)

2

= cos x+ 2nπ +3π

2

= cos(x+3π

= cos(x−π

= sin x

(d)

cos (x+kπ/2) = cos x+π(4n)

2

= cos (x+ 2nπ)

= cos x

14. (a) As the equation has a negative coeﬃcient, we need to shift this by a

half-period of πradians

y=−2 sin x

= 2 sin(x+π)

Comparing coeﬃcients, we get a= 2 and k= 1 and β=π

(b) As the equation has a negative coeﬃcient, we need to shift this by a

half-period of πradians

y= 2 sin(x−π/3)

= 2 sin(x−π/3 + π)

= 2 sin(x+ 2π/3)

Comparing coeﬃcients, we get a= 2 and k= 1 and β=−2π/3

half-period of πradians

y=−2 sin(x+π/4)

= 2 sin(x+π/4 + π)

= 2 sin(x+ 5π/4)

Comparing coeﬃcients, we get a= 2 and k= 1 and β=−5π/4

127

(d) As the equation is a cosine function, we need to re-express as a sine

function

3 cos x= 3 sin(x+π/2)

Comparing coeﬃcients, we get a= 3 and k= 1 and β=π/2

(e) As the equation is a cosine function, we need to re-express as a sine

function

y= 3 cos(x−π/6)

= 3 sin(x−π/6 + π/2)

= 3 sin(x+π/3)

Comparing coeﬃcients, we get a= 3 and k= 1 and β=−π/3

(f) As the equation is a cosine function, we need to re-express as a sine

function. Then re-express as a postive amplitude

y=−3 cos(x−π/6)

=−3 sin(x−π/6 + π/2)

= 3 sin(x−π/6 + π/2 + π)

= 3 sin(x+ 4π/3)

Comparing coeﬃcients, we get a= 3 and k= 1 and β=−4π/3

15.

cos(x−π/5)

1π2π3π4π

−1

16. We need to shift to the right by π

We need to shift to the left by 3π

128

17. To express kas an odd number, let k= 2n+ 1 for any integer n

tan (x+kπ/2) = tan x+π(2n+ 1)

2

= tan (x+ 2nπ +π/2)

= tan (x+π/2)

=sin(x+π/2)

cos(x+π/2)

=−cos x

sin x

=−cot x

To express kas an even number, let k= 2nfor any integer n

tan (x+kπ/2) = tan x+π(2n)

2

= tan (x+ 2nπ)

= tan x

Page 188

1. a) y= tan x−π

6

1π2π

−3

−2

−1

2π

b) y= 3 tan (x)

1π2π

−3

−2

−1

3π

c) y= cot x+π

4

129

1π2π

−3

−2

−1

3π

7π

2. Not really, although there is a trigonometric identity where

tan x= cot π

2−x

This equation has a negative xinside. Therefore it won’t be in form of

y= cot(x+ϕ) (note the positive x inside)

Page 189

1. (a) For y=y1+y2

1π2π

−4

−3

−2

−1

(b) For y=y1+y2

130

1π2π

−4

−3

−2

−1

1π2π

−4

−3

−2

−1

2. Graph 1a appear to be a sinusoidal function

Page 191

1. For y= 2 sin x+3 cos x, compare with standard form y=Asin kx+Bcos kx,

we get values of A= 2, B= 3, and k= 1

To convert to the other standard form y=asin k(x+ϕ), we calculate its

amplitude by

a=pA2+B2

=p22+ 32

=√13

131

While its phase ϕ=α/k is calculated by

cos α=A

√A2+B2

cos α=2

√13

α= arccos(2/√13)Since k= 1

ϕ= arccos(2/√13)

Therefore the equation is y=√2 sin (x+ϕ) where ϕ= arccos 2√13

Since amplitude is √2, then its maximum value is √2

2. Since amplitude is √13, then its maximum value is √13

3. For y= sin x+ cos x, compare with standard form y=Asin kx +Bcos kx,

we get values of A= 1, B= 1, and k= 1

To convert to the other standard form y=asin k(x+ϕ), we calculate its

amplitude aby

a=pA2+B2

=p12+ 12

=√2

While its phase ϕ=α/k is calculated by

cos α=A

√A2+B2

cos α=1

√2

α=π/4

Since k= 1

ϕ=π/4

Therefore the equation is y=√2sin (x+π/4)

Since amplitude is √2, then its maximum value is √2

4. For y= sin x−cos x, compare with standard form y=Asin kx +Bcos kx,

we get values of A= 1, B= 1, and k= 1

To convert to the other standard form y=asin k(x−β), we calculate its

amplitude by

a=pA2+B2

=p12+ 12

=√2

132

While its phase β=α/k is calculated by

cos α=A

√A2+B2

cos α=1

√2

α=π/4

Since k= 1

β=π/4

Therefore the equation is y=√2 sin (x−π/4)

Since amplitude is √2, then its maximum value is √2

5. For y= 4 sin x+3 cos x, compare with standard form y=Asin kx+Bcos kx,

we get values of A= 4, B= 3, and k= 1

To convert to the other standard form y=asin k(x+ϕ), we calculate its

amplitude by

a=pA2+B2

=p42+ 32

=√25

= 5

While its phase ϕ=α/k is calculated by

cos α=A

√A2+B2

cos α=4

α= arccos(4

Since k= 1

ϕ= arccos(4

Therefore the equation is y= 5 sin (x+ϕ) where ϕ= arccos(4/5)

Since amplitude is 5, then its maximum value is 5

6. For y= sin 2x+3 cos 2x, compare with standard form y=Asin kx+Bcos kx,

we get values of A= 1, B= 3, and k= 2

133

To convert to the other standard form y=asin k(x+ϕ), we calculate its

amplitude by

a=pA2+B2

=p12+ 32

=√10

While its phase ϕ=α/k is calculated by

cos α=A

√A2+B2

cos α=1

√10

α= arccos(1/√10)

Since k= 2

ϕ=1

2arccos( 1

√10)

Therefore the equation is y=√10 sin (x+ϕ) where ϕ=1

2arccos( 1

√10 )

7. For y= sin(x−π/4), compare with standard form y=asin k(x−β), we get

values of a= 1, β=π/4, and k= 1

To convert to the other standard form y=Asin kx −Bcos kx, we need to

solve for Aand Busing the deﬁnition A=acos kβ and B=asin kβ

A= 1 cos(π/4)

√2

B= 1 sin(π/4)

√2

Therefore the equation is y=1

√2sin x−1

√2icos x

8. For y= 4 sin(x+π/6), compare with standard form y=asin k(x+ϕ), we

get values of a= 4, ϕ=π/6, and k= 2

To convert to the other standard form y=Asin kx +Bcos kx, we need to

134

solve for Aand Busing the deﬁnition A=acos kβ and B=asin kβ

A= 4 cos(2π/6)

= 4 ×1/2

= 2

B= 4 sin(2π/6)

= 4 ×√3/2

= 2√3

Therefore the equation is y= 2 sin 2x+ 2√3 cos 2x

Page 194

1. (a)

y= 2 sin(x+π/6) + cos(x+π/6)

y= 2 (sin xcos π/6 + cos xsin π/6) + (cos xcos π/6−sin xsin π/6)

y= 2 √3

2sin x+1

2cos x!+ √3

2cos x−1

2sin x!

y=√3 sin x+ cos x+√3

2cos x−1

2sin x

y=√3−1

2sin x+ 1 + √3

2!cos x

(b)

y= 2 sin 2(x+π/4) −cos 2(x+π/4)

y= 2 sin(2x+π/2) −cos(2x+π/2)

y= 2 (sin 2xcos π/2 + cos 2xsin π/2) + (cos 2xcos π/2−sin 2xsin π/2)

y= 2 (sin 2x×0 + cos 2x×1) + (cos 2x×0−sin 2x×1)

y= 2 cos 2x−sin 2x

y=−sin 2x+ 2 cos 2x

2. (a)

y1+y1= 2 sin x+ sin(x−π/4)

= 2 sin x+ (sin xcos π/4−cos xsin π/4)

= 2 sin x+1

√2sin x−1

√2cos x

=2 + 1

√2sin x−1

√2cos x

(b) The functions have diﬀerent frequencies, which would need advanced

mathematics topics such as Fourier Series

135

Page 196

y=−x+ sin x

1π2π

−5

−4

−3

−2

−1

y=x2+ sin x

-π

y=x2+ cos x

Note this is an even function since f(−x) = f(x)

-π

136

y=x3+ sin x

-π

−5

−4

−3

−2

−1

y=x2+1

10 sin x

-π

y= cos x+1

10 sin 20x

−2π−1π1π2π

−2

−1

y= 2 sin x+1

10 sin 20x

137

−2π−1π1π2π

−2

−1

Page 197

1. In y= sin 2x+ sin 3x. The period of sin 2xhas form of m(2π/2), while the

period of sin 3xhas form of n(2π/3).

To be a period of both parts

m×2π

2=n×2π

3m= 2n

Using m= 2 and n= 3. the period of each part will be 2πradians

2. In y= sin 3x+ sin 6x. The period of sin 3xhas form of m(2π/3), while the

period of sin 6xhas form of n(2π/6).

To be a period of both parts

m×2π

3=n×2π

6m= 3n

2m= 1n

Using m= 1 and n= 2. the period of each part will be 2π/3 radians

3. In y= sin 4x+ sin 6x. The period of sin 4xhas form of m(2π/4), while the

period of sin 6xhas form of n(2π/6).

To be a period of both parts

m×2π

4=n×2π

6m= 4n

3m= 2n

Using m= 2 and n= 3. the period of each part will be πradians

5. In y= sin √2x+ sin 3√2x. The period of sin √2xhas form of m(2π/√2),

while the period of sin 3√2xhas form of n(2π/3√2).

138

To be a period of both parts

m×2π

√2=n×2π

3√2

3m= 1n

Using m= 1 and n= 3. the period of each part will be 2π/√2 = √2π

radians

Page 199

y= sin x−1

2sin 2x+1

3sin 3x

−2π−1π1π2π

−1

y= sin x−1

2sin 2x+1

3sin 3x−1

4sin 4x+1

5sin 5x

−2π−1π1π2π

−1

3. The limit will be a sawtooth wave

−2π−1π1π2π

−1

139

y= cos x−1

9cos 3x+1

25 cos 5x−1

49 cos 7x

−2π−1π1π2π

−1

Therefore, the limiting function is the triangular function

−2π−1π1π2π

−1

Page 200

1. The graph will be

1 2 3 4 5 6

2. The graph will be

1 2 3 4 5 6

−1

3. The graph will be

140

8 16 24 32

4. The graph will be

6 12 18 24

Tide is rising fastest at 3rd, 4th, 15th, and 16th hour

Tide is receding fastest at 9th, 10th, 21st, and 22nd hour

Tide is rising slowest at 1st, 5th, 13th and 17th hour

Tide is receding slowest at 7th, 11th, 19th and 23rd hour

Page 203

1. Substituing xvalues

•sin π= 0

•sin 2π= 0

•sin 3π= 0

•sin 4π= 0

2. For y= sin 4πx, the value of kis k= 4π, therefore the period is 2π

4π, so the

period is 1

3. Since period is 2π

k, if period is 3

3 = 2π

k=2π

141

Therefore the equation is y= sin 2π

4. Since period is 2π

k, if period is 2

2 = 2π

k=2π

Therefore the equation is y= sin πx

5. Since period is 2π

k, if period is n

n=2π

k=2π

Therefore the equation is y= sin 2π

Page 205

1. The period is estimated to be 12 months by distance between peaks at t= 8

and t= 20. Therefore the value for kis k= 2π/20

The amplitude is estimated to be mid-way between peak (15) and trough

(9.5), which is (15 −9.5)/2=2.75

Therefore the equation is y= 2.75 sin π

10 (x−π)

2. If (a) is in northern hemisphere, (b) and (c) would be in northern hemi-

sphere too as the peaks occur at similar month. (d) would be in southern

hemispehere as the wave is shifted by half-period

3. For all graphs, the ’average’ number of daylight hours occur at month 4,

month 10, month 16, and month 22

Chapter 9: Inverse Functions and Trigonometric

Equations

Page 213

1. (a) arcsin 0.5 = π/6

(b) arccos 0.5 = π/3

(d) arcsin(−√3/2) = −π/3

(e) arccos(−√3/2) = 5π/6

(f) arctan(−√3) = −π/3

142

(g) Note: not valid as the domain of arcsin xis −1≤x≤1

2. (a)

sin(arcsin 0.5)

= sin(π/6)

=−0.5

(b)

cos(arccos 0.5)

= cos(π/3)

= 0.5

(c)

tan(arctan(−1))

= tan(π/4)

=−π/4

(d)

arcsin(sin(π/3))

= arcsin(−√3/2)

=π/3

(e)

arccos(cos 11π/6)

= arccos(−√3/2)

=π/6

3. Let θ= arccos bwhich mean cos θ=b. If we assume angle θis acute, then

we can draw a right angle triangle

√1−b2

Giving sin θ=√1−b2when 0 ≤θ≤π/2

For π/2≤θπ, sin θ= sin(π−θ), making the sign positive

For π≤θ3π/2, sin θ=−sin(θ−π), making the sign negative

For 3π/2≤θ2π, sin θ=−sin(2θ−θ), making the sign negative

143

4. Let θ= arcsin bwhich mean sin θ=b. If we assume angle θis acute, then

we can draw a right angle triangle

√1−b2

Giving tan θ=b

√1−b2when 0 ≤θ≤π/2

For π/2≤θπ, tan θ=−tan(π−θ), making the sign negative

For π≤θ3π/2, tan θ= tan(θ−π), making the sign positive

For 3π/2≤θ2π, tan θ=−tan(2θ−θ), making the sign negative

5. Let θ= arctan bwhich mean tan θ=b. If we assume angle θis acute, then

we can draw a right angle triangle

√1 + b2

Giving cos θ=1

√1+b2when 0 ≤θ≤π/2

For π/2≤θπ, cos θ=−cos(π−θ), making the sign negative

For π≤θ3π/2, cos θ=−cos(θ−π), making the sign negative

For 3π/2≤θ2π, cos θ= cos(2θ−θ), making the sign positive

6. Since angle αis acute, then we can draw a right angle triangle

Note that sin α=a/b and cos β=a/b, and β=π/2−α

sin α= cos β

arccos(sin α) = β

arccos(sin α) = π/2−α

For π/2≤α≤π, the expression sin αbecomes sin(π−α), therefore making

arccos(sin α) = arccos(sin(π−α))

=π/2−(π/2−α)

=α

144

For π≤α≤3π/2, the expression sin αbecomes −sin(α−π), therefore

making

arccos(sin α) = arccos(−sin(α−π))

= arccos(sin(−α+π))

=π/2−(−α+π)

=α−π/2

For 3π/2≤α≤2π, the expression sin αbecomes −sin(2π−α), therefore

making

arccos(sin α) = arccos(−sin(2π−α))

= arccos(sin(−2π+α))

= arccos(sin(α))

=π/2−α

7. Note that the angles for parts (a) to (e) are less than π/2, so no change in

angle is needed, unlike (f)

(a) arcsin(sin π/11) = π/11

(b) arcsin(sin 2π/11) = 2π/11

(d) arcsin(sin 4π/11) = 4π/11

(e) arcsin(sin 5π/11) = 5π/11

(f)

arcsin(sin 6π/11)

= arcsin(sin(π−6π/11))

= arcsin(sin 5π/11)

= 5π/11

8. Graph of y= cos(arccos x)

145

−1 1

9. Graph of y= arccos(cos x)

−2π−1π1π2π

−1π

1π

10. Letting α= arcsin 3/5 and β= arcsin 5/13. Then drawing the triangles

146

sin(α+β) = sin αcos β+ cos αsin β

= 3/5×12/13 + 4/5×5/13

= 56/65

11. Letting α= arctan aand β= arctan b. Therefore we have tan α=aamd

tan β=b

tan(α+β) = tan α+ tan β

1−tan αtan β

=a+b

1−ab

12. Let tan α= 1/3 and tan β= 1/2 and tan γ= 1

tan(α+β) = tan α+ tan β

1−tan αtan β

tan(α+β) = 1/3+1/2

1−1/3×1/2

tan(α+β)=1

But Since both tan(α+β) = 1 and tan γ= 1. Therefore α+β=γ

Note: I think the author meant Question 11 instead of ’Problem 8’

13. Using plane geometry, draw another row of squares underneath the diagram

and draw lines AH and HC.

Note that the lengths of AH and HC are the same, making triangle AHC

an isosceles triangle, therefore CAH =ACH

Since triangle AHG and HEC is congruent with triangle BCD,BAH =β,

and HCE =β

Since triangle F CA is congruent with triangle ACD,F CA =α

Therefore

F CA +ACH +HCE = 90◦

α+ACH +β= 90◦

ACH = 90◦−α−β

Since γ= 45◦,CAH =α+βand CAH =ACH

α+β= 90◦−α−β

2(α+β) = 90◦

α+β= 45◦

α+β=γ

147

αβγ

A B

Note: I think the author meant Question 12 instead of ’Problem 9’

Page 220

1. From the graph, sin x > 1/2 when π/6≤x≤5π/6 and 13π/6≤x≤17π/6

2. Using the general equation πk + (−1)k(α) for any integer k

For sin x=−1/2, the value for αis arcsin(−1/2) = −π/6, so the solution is

πk + (−1)k(−π/6)

3. Using the general equation 2nπ ±(α) for any integer n

For cos x=√2/2, the value for αis arccos(√2/2) = π/4, so the solution is

2πn ±π/4

4. Using the general equation nπ + (α) for any integer n

For tan x= 1, the value for αis arctan(1) = π/4, so the solution is πn +π/4

5. Using the general equation πk + (−1)k(α) for any integer k

For sin x=−1, the value for αis arcsin(−1) = −π/2, so the solution is

πk + (−1)k(−π/2)

Page 221

1. Using the general equation πk + (−1)k(α) for any integer k

For sin x= sin π/5, the value for αis π/5, so the solution is πk + (−1)k(π/5)

2. Using the general equation πk + (−1)k(α) for any integer k

For sin x= sin π/2, the value for αis π/2, so the solution is πk + (−1)k(π/2)

148

1π2π3π4π5π6π

−1

cos α

α2π−α

Since period of cos xis 2π, the solutions are 2πn +αand 2πn −αfor any

integer n

4. Using the general equation 2nπ ±(α) for any integer n

For cos x= cos π/5, the value for αis π/5, so the solution is 2πn ±π/5

5. Since period of tan xis π, then if αis a solution, then the general equation

for solutions are α+πn for any integer n

1π2π

−3

−2

−1

3π

αα+π

6. Using the general equation nπ + (α) for any integer n

For tan x= tan π/5, the value for αis π/5, so the solution is πn +π/5

7. Using the general equation πk + (−1)k(α) for any integer k

sin x=−sin α

sin x= sin(−α)

x=πk + (−1)k(−α)

8. Using the general equation 2nπ ±(α) for any integer n

cos x=−cos α

cos x= cos(π−α)

x= 2nπ ±(π−α)

9. Using the general equation πk + (−1)k(α) for any integer k

sin x= sin α

sin x= sin(α)

x=πk + (−1)k(α)

149

Page 225

1. Using the general equation πk + (−1)k(α) for any integer k

sin 2x= 1

2x=πk + (−1)k×arcsin 1

2x=πk + (−1)k×π/2

x=1

2πk + (−1)k×π/2

2. Using the general equation πk + (−1)k(α) for any integer k

sin x/2=1/2

x/2 = πk + (−1)k×arcsin 1/2

x/2 = πk + (−1)k×π/6

x= 2 πk + (−1)k×π/6

cos x= sin 2x

cos x= 2 sin xcos x

0 = 2 sin xcos x−cos x

0 = cos x(2 sin x−1)

Therefore there are two set of solutions. Using general solutions 2nπ ±(α)

cos x= 0

x= 2nπ ±(arccos 0)

x= 2nπ ±(π/2)

and πk + (−1)k(α)

2 sin x−1 = 0

sin x= 1/2

x=πk + (−1)k(arcsin 1/2)

x=πk + (−1)k(π/6)

Note: I think the question is supposed to be cos x= sin2x?

150

sin x= sin 3x

sin x= sin(2x+x)

sin x= sin 2xcos x+ cos 2xsin x

sin x= 2 sin xcos xcos x+ (cos2x−sin2x) sin x

sin x= 2 sin x(1 −sin2x) + (1 −sin2x−sin2x) sin x

sin x= 2 sin x−2 sin3x+ sin x−2 sin3x

0 = −4 sin3x+ 2 sin x

0 = −2 sin x(2 sin2−1)

Therefore there are multiple sets of solutions to solve and using the general

equation πk + (−1)k(α)

sin x= 0

x=πk + (−1)k(arcsin 0)

x=πk

The other sets of solutions are

2 sin2x−1=0

sin2x= 1/2

sin x=±1/√2

For sin x= 1/√2

x=πk + (−1)k(arcsin 1/√2)

x=πk + (−1)k×π/4

For sin x=−1/√2

x=πk + (−1)k(arcsin −1/√2)

x=πk + (−1)k× −π/4

Note: I think the questions is supposed to be sin x= sin3x?

151

cos x= sin 4x

cos x= 2 sin 2xcos 2x

cos x= 2 ×2 sin xcos x×(cos2x−sin2x)

cos x= 4 sin xcos x×(1 −sin2x−sin2x)

0 = 4 sin xcos x×(1 −2 sin2x)−cos x

0 = cos x4 sin x(1 −2 sin2x)−1

0 = cos x−8 sin3x+ 4 sin x−1

Therefore there are multiple sets of solutions to solve and using the general

equation πk + (−1)k(α) and 2nπ ±(α)

cos x= 0

x= 2nπ ±arccos 0

x= 2nπ ±π/2

Solving −8 sin3x+ 4 sin x−1 = 0 requires using cubic functions, which is

advanced level and is beyond this book’s level

Note: I think the question is supposed to be cos x= sin 4x?

26 sin2x+ cos2x= 10

26 sin2x+ 1 −sin2x−10 = 0

25 sin2x−9=0

(5 sin x+ 3)(5 sin x−3) = 0

Therefore there are multiple sets of solutions to solve and using the general

equation πk + (−1)k(α)

For 5 sin x+ 3 = 0

5 sin x+ 3 = 0

sin x=−3/5

x=πk + (−1)k(arcsin −3/5)

For 5 sin x−3=0

5 sin x−3=0

sin x= 3/5

x=πk + (−1)k(arcsin 3/5)

152

cos2x−cos x= sin2x

cos2x−cos x= 1 −cos2x

2 cos2x−cos x−1=0

(2 cos x+ 1)(cos x−1) = 0

Therefore there are multiple sets of solutions to solve and using the general

equation 2πn ±(α)

For 2 cos x+ 1 = 0

2 cos x+ 1 = 0

cos x=−1/2

x= 2πn ±arccos(−1/2)

x= 2πn ±2π/3

For 2 cos x−1=0

2 cos x−1=0

cos x= 1/2

x= 2πn ±arccos(1/2)

x= 2πn ±π/3

3 tan2x= 12

tan2x= 4

tan2x−4=0

(tan x+ 2)(tan x−2) = 0

Therefore there are multiple sets of solutions to solve and using the general

equation πk + (α)

For tan x+ 2 = 0

tan x+ 2 = 0

tan x=−2

x=πk + arctan −2

For tan x−2=0

tan x−2=0

tan x= 2

x=πk + arctan 2

153

cos 2x= 2 sin2x

cos2x−sin2x= 2 sin2x

1−sin2x−sin2x= 2 sin2x

1−4 sin2x= 0

(1 −2 sin x)(1 + 2 sin x) = 0

Therefore there are multiple sets of solutions to solve and using the general

equation πk + (−1)k(α)

For 1 −2 sin x= 0

1−2 sin x= 0

sin x= 1/2

x=πk + (−1)karcsin 1/2

x=πk + (−1)kπ/6

For 1 + 2 sin x= 0

1 + 2 sin x= 0

sin x=−1/2

x=πk + (−1)karcsin −1/2

x=πk + (−1)k(−π/6)

10.

tan2x= cot x

tan3x= 1

tan3x−1 = 0

(tan x−1)(tan2x+ tan x+ 1) = 0

Therefore there are multiple sets of solutions to solve and using the general

equation πk +α

For tan x−1=0

tan x−1=0

tan x=1

x=πk + arctan 1

x=πk +π/4

154

For tan2x+ tan x+ 1 = 0, apply quadratic formula

tan x=−1±p12−4(1)(1)

2(1)

tan x=−1±√−3

Which does not have any solutiosn as the number inside the square root is

negative

11.

cos2x= 7 tan x+ 3

5 sec2x= 7 tan x+ 3

5(tan2x+ 1) = 7 tan x+ 3

5 tan2x+ 5 = 7 tan x+ 3

5 tan2x−7 tan x+ 2 = 0

Apply quadratic formula

tan x=7±p49 −4(5)(2)

2(5)

tan x=7±3

Therefore there are multiple sets of solutions to solve and using the general

equation πk +α

For tan x=7+3

tan x= 1

x=πk + arctan 1

x=πk +π/4

For tan x=7−3

tan x= 2/5

x=πk + arctan 2/5

12.

√3 tan2x+ 1 = (1 + √3) tan x

(√3 tan x+ 1)(tan x+ 1) = 0

155

Therefore there are multiple sets of solutions to solve and using the general

equation πk +α

For √3 tan x+ 1 = 0

√3 tan x+ 1 = 0

√3 tan x=−1

tan x=−1/√3

x=πn + arctan −1/√3

x=πn + (−π/6)

For tan x+ 1 = 0

tan x+ 1 = 0

tan x=−1

x=πn + arctan −1

x=πn + (−π/4)

13. For Solution 1, re-express the equations such that

with n

x=π/6+2πn/3

with n+1

=π/6+2π(n+ 1)/3

=π/6+2πn/3+2π/3

= 5π/6+2πn/3

with n+2

=π/6+2π(n+ 2)/3

=π/6+2πn/3+4π/3

= 9π/6+2πn/3

= 3π/2+2πn/3

Which corresponds to each line of Solution 2, provided that the number n for

Solution 1 is multiples of 3

Page 227

1. If α= 1.6 then

x=α−tan α

≈35.8325

156

Since near x=π/2, the tangent will be very ﬂat and only reaches the x-axis

after travelling a long distance

2. If α=π/4 then

x=α−tan α

≈ −0.2146

Since the tangent is intersecting x axis to the left

Note, the smaller the α, the closer to it intersecting x-axis at x= 0

If α= 0.1 then

x=α−tan α

≈ −0.000334

3. To solve for location of R where x= 0 and noting x≈tan xfor small x

value

x=α−tan α

0 = α−tan α

tan α=α

α= 0

Therefore R will be at origin when we take a tangent at x= 0

4. if x=π/2, then the tangent will be horizontal, therefore not intersecting

with x-axis at all

If x is a bit smaller than π/2, then the tangent will intersect with x-axis to

the left (far away)

If x is a bit bigger than π/2, then the tangent will intersect with x-axis to

the right (far away)

Page 229

1. Using A≈h

sin h/2, where h=π/n

Using n= 4, so h=π/4 gives A≈π/4

sin π/8, which is 2.05234

Using n= 8, so h=π/8 gives A≈π/8

sin π/16 , which is 2.01291

2. Since area between 0 ≤x≤πis approximately 2, threfore the area under

curve y= sin xfrom x= 0 to x=π/2 would be around 1

157

hsin x1+hsin x2+hsin x3+··· +hsin xm

=hsin π

n+ sin 2π

n+··· + sin nπ

2π

=hsin n+1

2nsin n

sin π

=hsin n+1

sin π

=hsin n+1

sin π

√2

if n is very large

≈hsin π/4

sin π/2n

√2

≈h

2 sin h/2

letting n= 8, giving h=π/8, gives approximate area of π/8

2 sin π/16 ≈1.00645

158