Solution Guide for Chapter 8: Relationships Among...

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Solution Guide for Chapter 8: Relationships Among Trigonometric Functions 8.1 TRIGONOMETRIC IDENTITIES E-1. Half-angle formula for the cosine: By the Pythagorean theorem applied to the triangle in Figure 8.14, x 2 + 1 - cos α 2 2 =1. Hence x 2 =1 - 1 - cos α 2 = 1 + cos α 2 . Now we take the square root: x = 1 + cos α 2 . Because x = cos α 2 , this gives the formula cos α 2 = 1 + cos α 2 . E-2. Half-angle formula for the sine: We modify the triangle in Figure 8.14, giving the label x = sin α 2 to the side opposite the angle α 2 and the label cos α 2 = 1 + cos α 2 to the side adjacent to that angle. By the Pythagorean theorem applied to the triangle, x 2 + 1 + cos α 2 2 =1. Hence x 2 =1 - 1 + cos α 2 = 1 - cos α 2 . Now we take the square root: x = 1 - cos α 2 . Because x = sin α 2 , this gives the formula sin α 2 = 1 - cos α 2 . E-3. Halving, then doubling angles: As suggested, first we use the double-angle formula for the sine, then we use the half-angle formulas: sin 2 α 2 = 2 sin α 2 cos α 2 =2 1 - cos α 2 1 + cos α 2 =2 1 - cos 2 α 4 .

Transcript of Solution Guide for Chapter 8: Relationships Among...

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Solution Guide for Chapter 8:Relationships AmongTrigonometric Functions

8.1 TRIGONOMETRIC IDENTITIES

E-1. Half-angle formula for the cosine: By the Pythagorean theorem applied to the triangle

in Figure 8.14,

x2 +

(√1− cos α

2

)2

= 1.

Hence

x2 = 1− 1− cos α

2=

1 + cos α

2.

Now we take the square root: x =√

1 + cos α

2. Because x = cos

α

2, this gives the formula

cosα

2=

√1 + cos α

2.

E-2. Half-angle formula for the sine: We modify the triangle in Figure 8.14, giving the label

x = sinα

2to the side opposite the angle

α

2and the label cos

α

2=

√1 + cos α

2to the side

adjacent to that angle. By the Pythagorean theorem applied to the triangle,

x2 +

(√1 + cos α

2

)2

= 1.

Hence

x2 = 1− 1 + cos α

2=

1− cos α

2.

Now we take the square root: x =√

1− cos α

2. Because x = sin

α

2, this gives the formula

sinα

2=

√1− cos α

2.

E-3. Halving, then doubling angles: As suggested, first we use the double-angle formula

for the sine, then we use the half-angle formulas:

sin(2(α

2

))= 2 sin

2

)cos(α

2

)= 2

(√1− cos α

2

)(√1 + cos α

2

)= 2

√1− cos2 α

4.

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SECTION 8.1 Trigonometric Identities 571

But 1− cos2 α = sin2 α, so

2

√1− cos2 α

4= 2

√sin2 α

4= sinα.

E-4. Doubling, then halving angles: As suggested, first we use the half-angle formula for

the sine:

sin(

2

)=

√1− cos(2α)

2.

Now we use the double-angle formula for the cosine and the most basic trig identity:

1− cos(2α) = 1− (cos2 α− sin2 α) = 1− cos2 α + sin2 α = sin2 α + sin2 α = 2 sin2 α.

Thus

sin(

2

)=

√1− cos(2α)

2=

√2 sin2 α

2= sinα.

S-1. The sine of 75 degrees: Now

sin 75◦ = sin(45◦ + 30◦) = sin 45◦ cos 30◦ + cos 45◦ sin 30◦ =1√2×√

32

+1√2× 1

2= 0.97.

S-2. The cosine of 75 degrees: Now

cos 75◦ = cos(45◦ + 30◦) = cos 45◦ cos 30◦− sin 45◦ sin 30◦ =1√2×√

32− 1√

2× 1

2= 0.26.

S-3. The sine of 15 degrees: Now

sin 15◦ = sin(45◦ − 30◦) = sin 45◦ cos 30◦ − cos 45◦ sin 30◦ =1√2×√

32− 1√

2× 1

2= 0.26.

S-4. The cosine of 15 degrees: Now

cos 15◦ = cos(45◦− 30◦) = cos 45◦ cos 30◦ + sin 45◦ sin 30◦ =1√2×√

32

+1√2× 1

2= 0.97.

S-5. Half angle: Now

sin 22.5◦ = sin(

1245◦)

=

√1− cos 45◦

2=

√1− 1√

2

2= 0.38.

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572 Solution Guide for Chapter 8

S-6. Half angle: Now

cos 22.5◦ = cos(

1245◦)

=

√1 + cos 45◦

2=

√1 + 1√

2

2= 0.92.

S-7. Half angle twice: Now

sin 11.25◦ = sin(

1222.5◦

)=

√1− cos 22.5◦

2=

√√√√√1−√

1 + cos 45◦

22

=

√√√√√1−√

1 + 1/√

22

2= 0.20.

S-8. Double angle: Now

sin 120◦ = sin (2× 60◦) = 2 sin 60◦ cos 60◦ = 2×√

32× 1

2=√

32

= 0.87.

S-9. Double-angle formula: Now

cos 120◦ = cos(2× 60◦) = cos2 60◦ − sin2 60◦ =(

12

)2

(√3

2

)2

= −12.

S-10. Sum formula: Now

sin 105◦ = sin(45◦+60◦) = sin 45◦ cos 60◦+cos 45◦ sin 60◦ =1√2× 1

2+

1√2×√

32

= 0.97.

S-11. Finding angles:

(a) Because t is between 0 and 90 degrees, cos t is positive. Hence

cos t =√

1− sin2 t =√

1− 0.72 = 0.71.

(b) Using the double-angle formula for cosine, along with the most basic trig identity,

gives

cos(2t) = cos2 t− sin2 t = (1− sin2 t)− sin2 t = (1− 0.72)− 0.72 = 0.02.

Using the rounded value of cos t from Part (a) gives the value 0.01 here.

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SECTION 8.1 Trigonometric Identities 573

(c) Because t is between 0 and 90 degrees,t

2is between 0 and 45 degrees, so sin

t

2is

positive. Using the half-angle formula for sine along with Part (a) gives

sint

2=

√1− cos t

2= 0.38.

(d) Using the double-angle formula for sine twice, along with Part (a) and Part (b),

gives

sin(4t) = 2 sin(2t) cos(2t) = 2(2 sin t cos t) cos(2t) = 0.04.

The answer will vary depending on rounding in preceding parts.

S-12. Finding trigonometric functions of angles:

(a) Because t is between 90 and 180 degrees, cos t is negative. Hence

cos t = −√

1− sin2 t = −√

1− 0.62 = −0.80.

(b) Using the double-angle formula for cosine, along with the most basic trig identity,

gives

cos(2t) = cos2 t− sin2 t = (1− sin2 t)− sin2 t = (1− 0.62)− 0.62 = 0.28.

(c) Because t is between 90 and 180 degrees,t

2is between 45 and 90 degrees, so sin

t

2is positive. Using the half-angle formula for sine along with Part (a) gives

sint

2=

√1− cos t

2= 0.95.

(d) Using the double-angle formula for sine twice, along with Part (a) and Part (b),

gives

sin(4t) = 2 sin(2t) cos(2t) = 2(2 sin t cos t) cos(2t) = −0.54.

S-13. Calculating sines of angles:

(a) Now sin(−30◦) = − sin 30◦ = −12.

(b) Now cos(−60◦) = cos 60◦ =12.

(c) By the sum formula for the tangent in Example 8.2,

tan 105◦ = tan(60◦ + 45◦) =tan 60◦ + tan 45◦

1− tan 60◦ tan 45◦.

But

tan 60◦ =sin 60◦

cos 60◦=

√3

212

=√

3

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574 Solution Guide for Chapter 8

and

tan 45◦ =sin 45◦

cos 45◦=

1√2

1√2

= 1.

Hence

tan 105◦ =tan 60◦ + tan 45◦

1− tan 60◦ tan 45◦=√

3 + 11−

√3

= −3.73.

S-14. Area: The area is12× 5× 6× sin 30◦ =

12× 5× 6× 1

2= 7.5.

S-15. Area: The area is12× 5× 8× sin 25◦ = 8.45.

S-16. Going backward: We want to find t so that sin t = 0.78. We use the crossing-graphs

method with a horizontal span from 0 to 90 using degrees and a vertical span from 0 to

1. We see from the graph below that t = 51.26 degrees.

S-17. Going backward: We want to find t so that cos t = 0.45. We use the crossing-graphs

method with a horizontal span from 0 to 90 using degrees and a vertical span from 0 to

1. We see from the graph below that t = 63.26 degrees.

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SECTION 8.1 Trigonometric Identities 575

S-18. Going backward: We want to find t so that cos t = −0.34. We use the crossing-graphs

method with a horizontal span from 0 to 180 using degrees and a vertical span from −1

to 1. We see from the graph below that t = 109.88 degrees.

1. Unit circle: According to the unit-circle definition of sine and cosine, in magnitude sin t

and cos t are the legs of a right triangle whose hypotenuse is 1 (the radius of the unit

circle). By the Pythagorean theorem, sin2 t + cos2 t = 1.

2. Sum formula for the cosine:

(a) Because sine is odd, sin(x− 90◦) = − sin(90◦ − x) = − cos x.

(b) Because cosine is even, cos(x− 90◦) = cos(90◦ − x) = sin x.

(c) Now

cos(α + β) = − sin(α + β − 90◦) (by Part (a))

= − sin((α− 90◦) + β)

= − sin(α− 90◦) cos β − cos(α− 90◦) sinβ (by the sum formula)

= cos α cos β − sinα sinβ (by Part (a) and Part (b)).

3. Double angle for the cosine: Using the sum formula for the cosine gives

cos(2t) = cos(t + t) = cos t cos t− sin t sin t = cos2 t− sin2 t.

4. Difference formula for the sine: Using the sum formula for the sine gives

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

because sine is odd and cosine is even.

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576 Solution Guide for Chapter 8

5. Difference formula for the cosine: Using the sum formula for the cosine gives

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

because sine is odd and cosine is even.

6. An identity:

(a) We use the double-angle formulas:

cot(2x) =cos(2x)sin(2x)

=cos2 x− sin2 x

2 sinx cos x=

cos2 x

2 sinx cos x− sin2 x

2 sinx cos x=

cos x

2 sinx− sinx

2 cos x

=12(cot x− tanx).

(b) We start with the right-hand side:

cot2 x− 12 cot x

=12

(cot2 x

cot x− 1

cot x

)=

12(cot x− tanx) = cot(2x)

by Part (a).

7. Changing how things look: Using the double-angle formula for cosine, along with the

most basic trig identity, gives

cos(2x) = cos2 x− sin2 x = cos2 x− (1− cos2 x) = 2 cos2 x− 1.

Hence 2 cos2 x = 1 + cos(2x), so cos2 x =12(1 + cos(2x)).

8. Changing how things look: As suggested, we start with the right-hand side of the

equation, and we apply the sum formula and the difference formula for the sine:

12(sin(x + y) + sin(x− y)) =

12(sinx cos y + cos x sin y + sinx cos y − cos x sin y) =

12(2 sinx cos y)

= sinx cos y.

9. An important identity:

(a) By the sum formula for the cosine,

cos 3t = cos(2t + t) = cos 2t cos t− sin 2t sin t.

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SECTION 8.1 Trigonometric Identities 577

(b) Using the double-angle formulas in Part (a) gives

cos 3t = (cos2 t− sin2 t)(cos t)− (2 sin t cos t)(sin t) = cos3 t− 3 cos t sin2 t.

If we replace sin2 t by 1− cos2 t, this becomes

cos3 t− 3 cos t(1− cos2 t) = 4 cos3 t− 3 cos t.

(c) By Part (b) we have

4 cos3 20◦ − 3 cos 20◦ = cos(3(20◦)) = cos(60◦) =12,

so

4 cos3 20◦ − 3 cos 20◦ − 12

= 0.

Hence cos 20◦ is a zero of the polynomial 4x3 − 3x− 12.

10. Double-angle formula for the tangent: By the sum formula for the tangent we have

tan(2x) = tan(x + x) =tanx + tanx

1− tanx tanx=

2 tanx

1− tan2 x.

11. Sum formula for the cotangent: As in Example 8.2 we have

tan(α + β) =sin(α + β)cos(α + β)

=sinα cos β + cos α sinβ

cos α cos β − sinα sinβ.

Now we divide each term of the top and bottom by sinα sinβ to get

tan(α + β) =sin α cos βsin α sin β + cos α sin β

sin α sin β

cos α cos βsin α sin β − sin α sin β

sin α sin β

.

Cancelling gives

tan(α + β) =cos βsin β + cos α

sin α(cos αsin α

) (cos βsin β

)− 1

=cot α + cot β

cot α cot β − 1.

But then

cot(α + β) =1

tan(α + β)=

cot α cot β − 1cot α + cot β

.

12. Double-angle formula for the cotangent again: By Exercise 11 we have

cot(2x) = cot(x + x) =cot x cot x− 1cot x + cot x

=cot2 x− 1

2 cot x.

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578 Solution Guide for Chapter 8

13. Area of an isosceles triangle:

(a) Now cotangent is adjacent divided by opposite, so cotα

2=

Ha2

. Solving for H gives

H =a

2cot

α

2.

(b) Now

Area =12× Length of base × Length of altitude =

12× a×H.

Using the formula for H from Part (a) gives

Area =12× a× a

2cot

α

2=

14a2 cot

α

2.

14. Area of a regular polygon: As suggested, we divide the polygon into n isosceles trian-

gles, each with angle of360n

degrees opposite a base of length a. By Exercise 13 the area

of each triangle is14a2 cot

360◦

2n=

14a2 cot

180◦

n.

There are n such triangles, so the area of the polygon is14na2 cot

180◦

n.

8.2 LAWS OF SINES AND COSINES

E-1. Thales revisited: For an angle inscribed in a semicircle the degree measure of the central

angle is 180◦. Thus the measure of the inscribed angle is half of 180◦, so it is 90◦. Thus

the inscribed angle is a right angle.

E-2. Converse: Given a right triangle, form the circumscribed circle. The hypotenuse is

a chord of that circle, and the angle subtended by the chord is the right angle of the

triangle, so the subtended angle has measure 90◦. Because the degree measure of the

subtended angle is half the degree measure of the central angle, the central angle is

2× 90◦ = 180◦. Hence the chord is a diameter of the circle.

E-3. Special value:

(a) Because the degree measure of any subtended angle is half the degree measure of

the central angle, the degree measure of β is half the degree measure of α. Now the

measure of β is 30◦, so the measure of α is 60◦.

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SECTION 8.2 Laws of Sines and Cosines 579

(b) We consider the angles of the triangle ∆AOB, namely OAB, OBA, and α. Now

OB and OA are radii, so they have the same length. Thus the angles opposite these

sides, namely OAB and OBA, are equal. By Part (a) the measure of α is 60◦. Now

the angle sum of ∆AOB is 180 degrees, and because OAB and OBA are equal the

measure of each must be12(180 − 60) = 60 degrees. Hence all angles of ∆AOB

must be 60◦, so the triangle is equilateral.

(c) By Part (b), the side AB has the same length as the side OA, and the latter is a

radius. Hence the length of AB equals the radius.

(d) Now the length of AB divided by sinβ equals the diameter, and it follows that sinβ

is the length of AB divided by the diameter. But by Part (c) the length of AB equals

the radius, so it equals half the diameter. Thus sinβ is half the diameter divided by

the diameter. Because the measure of β is 30◦, this says that sin 30◦ =12.

S-1. First we find a: Because the angle sum of the triangle is 180 degrees, a = 180−40−80 =

60 degrees. Now we use the law of sines to find B:

B

sin 40◦=

12sin 60◦

,

so

B = sin 40◦ × 12sin 60◦

= 8.91.

Similarly,C

sin 80◦=

12sin 60◦

,

so

C = sin 80◦ × 12sin 60◦

= 13.65.

S-2. First we find a: Because the angle sum of the triangle is π radians, a = π − π

6− π

4=

12= 1.83 radians. Now we use the law of sines to find B:

B

sin π6

=12

sin 7π12

,

so

B = sinπ

6× 12

sin 7π12

= 6.21.

Similarly,C

sin π4

=12

sin 7π12

,

so

C = sinπ

4× 12

sin 7π12

= 8.78.

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580 Solution Guide for Chapter 8

S-3. First we find b: Because the angle sum of the triangle is 180 degrees, b = 180−55−65 = 60

degrees. Now we use the law of sines to find A:

A

sin 55◦=

10sin 60◦

,

so

A = sin 55◦ × 10sin 60◦

= 9.46.

Similarly,C

sin 65◦=

10sin 60◦

,

so

C = sin 65◦ × 10sin 60◦

= 10.47.

S-4. First we find b: Because the angle sum of the triangle is π radians, b = π−1.5−0.2 = 1.44

radians. Now we use the law of sines to find A:

A

sin 1.5=

7sin 1.44

,

so

A = sin 1.5× 7sin 1.44

= 7.04.

Similarly,C

sin 0.2=

7sin 1.44

,

so

C = sin 0.2× 7sin 1.44

= 1.40.

S-5. First we use the law of cosines to find C:

C2 = 82 + 52 − 2× 8× 5 cos 30◦,

so

C =√

82 + 52 − 2× 8× 5 cos 30◦ = 4.44.

Now we use the law of sines to find b (the angle opposite B, the smaller of the two

remaining sides):5

sin b=

4.44sin 30◦

,

so

sin b = 5× sin 30◦

4.44.

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SECTION 8.2 Laws of Sines and Cosines 581

Using the crossing-graphs method gives b = 34.27 degrees. (There are two solutions

between 0 and 180 degrees, but, because B isn’t the longest side of the triangle, the

angle b opposite it must be less than 90 degrees.) Because the angle sum of the triangle

is 180 degrees, a = 180− 34.27− 30 = 115.73 degrees.

S-6. First we use the law of cosines to find C:

C2 = 82 + 52 − 2× 8× 5 cosπ

7,

so

C =√

82 + 52 − 2× 8× 5 cosπ

7= 4.11.

Now we use the law of sines to find b (the angle opposite B, the smaller of the two

remaining sides):5

sin b=

4.11sin π

7

,

so

sin b = 5×sin π

7

4.11.

Using the crossing-graphs method gives b = 0.56 radian. (There are two solutions be-

tween 0 and π radians, but, because B isn’t the longest side of the triangle, the angle

b opposite it must be less thanπ

2radians.) Because the angle sum of the triangle is π

radians, a = π − 0.56− π

7= 2.13 radians.

S-7. First we use the law of sines to find c:

5sin c

=10

sin 110◦,

so

sin c = 5× sin 110◦

10.

Using the crossing-graphs method gives c = 28.02 degrees. (There are two solutions

between 0 and 180 degrees, but, because b is greater than 90 degrees, c must be less

than 90 degrees.) Now we find a: Because the angle sum of the triangle is 180 degrees,

a = 180− 110− 28.02 = 41.98 degrees. Now we use the law of sines to find A:

A

sin 41.98◦=

10sin 110◦

,

so

A = sin 41.98◦ × 10sin 110◦

= 7.12.

(We can also find A using the law of cosines.)

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582 Solution Guide for Chapter 8

S-8. First we use the law of sines to find c:

5.5sin c

=10

sin 2.3,

so

sin c = 5.5× sin 2.310

.

Using the crossing-graphs method gives c = 0.42 radian. (There are two solutions be-

tween 0 and π radians, but, because a is greater thanπ

2radians, c must be less than

π

2radians.) Now we find b: Because the angle sum of the triangle is π radians, b =

π − 2.3− 0.42 = 0.42 radian. Now we use the law of sines to find B:

B

sin 0.42=

10sin 2.3

,

so

B = sin 0.42× 10sin 2.3

= 5.47.

(We can also find B using the law of cosines.) Note that, because b = c, we expect that

B = C. The difference is due to rounding.

S-9. First we find c: Because the angle sum of the triangle is 180 degrees, c = 180−30−120 =

30 degrees. Because a and c are equal, A and C are equal, so C = 10. (We can also find

C using the law of sines.) Now we use the law of sines to find B:

B

sin 120◦=

10sin 30◦

,

so

B = sin 120◦ × 10sin 30◦

.

Now sin 120◦ = sin 60◦ =√

32

because the reference angle for 120◦ is 180 − 120 = 60

degrees. Also, sin 30◦ =12

. Hence

B =√

32× 10

12

= 10√

3.

(We can also find B using the law of cosines.)

S-10. First we find c: Because the angle sum of the triangle is 180 degrees, c = 180−60−60 = 60

degrees. Because a, b, and c are all equal, this is an equilateral triangle. Hence A, B, and

C are all equal, so B = 5 and C = 5. (We can also find B and C using the law of sines.)

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SECTION 8.2 Laws of Sines and Cosines 583

1. Finding angles: First we use the law of cosines to find c because C is the longest side:

102 = 72 + 92 − 2× 7× 9 cos c,

so

cos c =72 + 92 − 102

2× 7× 9.

Using the crossing-graphs method gives c = 76.23 degrees. (The horizontal span is 0 to

180 using degrees.) Now we use the law of sines to find a:

7sin a

=10

sin 76.23◦,

so

sin a = 7× sin 76.23◦

10.

Using the crossing-graphs method gives a = 42.83 degrees. (There are two solutions

between 0 and 180 degrees, but, because A isn’t the longest side of the triangle, the

angle a opposite it must be less than 90 degrees.) Because the angle sum of the triangle

is 180 degrees, b = 180− 42.83− 76.23 = 60.94 degrees.

2. Finding distance: First we find the angle at city 3: Because the angle sum of the triangle

is 180 degrees, the angle at city 3 is 180− 80− 70 = 30 degrees. Now we use the law of

sines to find the distance in miles from city 1 to city 3, which we denote by A:

A

sin 70◦=

200sin 30◦

,

so

A = sin 70◦ × 200sin 30◦

= 375.88.

Thus the distance from city 1 to city 3 is about 376 miles. Similarly, if B denotes the

distance (in miles) from city 2 to city 3, then

B

sin 80◦=

200sin 30◦

,

so

B = sin 80◦ × 200sin 30◦

= 393.92.

Thus the distance from city 2 to city 3 is about 394 miles.

3. Mapping a mountain: We use the law of cosines: If C denotes the length (in yards) of

the side of the hill, then

C2 = 3002 + 5002 − 2× 300× 500 cos 20◦,

so

C =√

3002 + 5002 − 2× 300× 500 cos 20◦ = 241.02.

Thus the length is about 241 yards.

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584 Solution Guide for Chapter 8

4. A lost airplane: The two airports and the airplane determine a triangle. The angle at

OKC is 90−44 = 46 degrees, and the angle at DFW is 90−62 = 28 degrees. Because the

angle sum of the triangle is 180 degrees, the angle at the airplane is 180− 46− 28 = 106

degrees. Now we use the law of sines to find the distance in miles from DFW to the

airplane, which we denote by A:

A

sin 46◦=

200sin 106◦

,

so

A = sin 46◦ × 200sin 106◦

= 149.67.

Thus the distance from DFW to the airplane is about 150 miles. Similarly, if B denotes

the distance (in miles) from OKC to the airplane, then

B

sin 28◦=

200sin 106◦

,

so

B = sin 28◦ × 200sin 106◦

= 97.68.

Thus the distance from OKC to the airplane is about 98 miles.

5. The base of an isosceles triangle:

(a) By the law of cosines we have

C2 = r2 + r2 − 2× r × r × cos γ = 2r2(1− cos γ) = 4r2

(1− cos γ

2

).

(b) Taking the square root in Part (a) gives

C = 2r

√1− cos γ

2,

and by the half-angle formula for sine this says

C = 2r sinγ

2.

6. An inscribed polygon:

(a) As suggested, we divide the polygon into n isosceles triangles, each of which is

described by Figure 8.33 with γ =360◦

n. By Part (b) of Exercise 5 we have

C = 2r sinγ

2= 2r sin

180◦

n.

There are n such triangles, so the perimeter of the polygon is

nC = 2nr sin180◦

n.

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SECTION 8.2 Laws of Sines and Cosines 585

(b) The limiting value, as the number of sides increases, of the perimeters of inscribed

polygons should be the circumference of the circle, namely 2πr. In fact, if we put

r = 1 (say) in the formula from Part (a) and examine a table of values for the

resulting function of n, namely

2n sin180◦

n,

we see that the limiting value is about 6.28, an approximation of 2π.

7. A baseball field: The center fielder, home plate, and third base determine a triangle.

The angle of this triangle at home plate is 45 degrees because the side of the triangle

from home plate to the center fielder includes a diagonal of the square shown in the

figure. If C denotes the distance (in feet) from the center fielder to third base, then by

the law of cosines

C2 = 3802 + 902 − 2× 380× 90 cos 45◦,

so

C =√

3802 + 902 − 2× 380× 90 cos 45◦ = 322.70.

Thus the distance is about 323 feet.

8. An identity:

(a) Now

S−C =12(A+B+C)−C =

12(A+B)+

12C−C =

12(A+B)− 1

2C =

12(A+B−C).

(b) Now by the definition of S and Part (a)

S(S−C) =12(A+B+C)×1

2(A+B−C) =

14((A+B)+C)((A+B)−C) =

14((A+B)2−C2).

Here we used the identity from algebra

(x + y)(x− y) = x2 − y2.

Hence

S(S−C) =14((A+B)2−C2) =

14((A2+2AB+B2)−C2) =

14(A2+B2+2AB−C2).

Here we used the identity from algebra

(x + y)2 = x2 + 2xy + y2.

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586 Solution Guide for Chapter 8

(c) Now the law of cosines gives

C2 = A2 + B2 − 2AB cos γ.

Substituting this into the result from Part (b) gives

S(S − C) =14(A2 + B2 + 2AB − C2)

=14(A2 + B2 + 2AB − (A2 + B2 − 2AB cos γ))

=14(2AB + 2AB cos γ)

=12AB(1 + cos γ).

(d) By Part (c) and the half-angle formula for the cosine,

S(S − C) =12AB(1 + cos γ) = AB cos2

γ

2,

soS(S − C)

AB= cos2

γ

2.

Taking the square root gives √S(S − C)

AB= cos

γ

2.

(Note that γ is between 0 and 180 degrees, soγ

2is between 0 and 90 degrees, and

thus cosγ

2is nonnegative.)

9. Another identity:

(a) Now

S−A =12(A+B+C)−A =

12A+

12(B+C)−A = −1

2A+

12(B+C) =

12(C+B−A).

This can also be written as12(C + (B −A)).

A similar computation gives

S −B =12(C −B + A) =

12(C − (B −A)).

Hence

(S −A)(S −B) =12(C + (B −A))

12(C − (B −A)).

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SECTION 8.2 Laws of Sines and Cosines 587

(b) We use Part (a) along with the following identity from algebra:

(x + y)(x− y) = x2 − y2.

We get

(S −A)(S −B) =12(C + (B −A))

12(C − (B −A)) =

14(C2 − (B −A)2).

Expanding using the identity

(x− y)2 = x2 − 2xy + y2

gives

(S−A)(S−B) =14(C2−(B−A)2) =

14(C2−(B2−2AB+A2)) =

14(C2−B2−A2+2AB).

(c) Now the law of cosines gives

C2 = A2 + B2 − 2AB cos γ.

Substituting this into the result from Part (b) gives

(S −A)(S −B) =14(C2 −B2 −A2 + 2AB)

=14(A2 + B2 − 2AB cos γ −B2 −A2 + 2AB)

=14(2AB − 2AB cos γ)

=12AB(1− cos γ).

(d) By Part (c) and the half-angle formula for the sine,

(S −A)(S −B) =12AB(1− cos γ) = AB sin2 γ

2,

so(S −A)(S −B)

AB= sin2 γ

2.

Taking the square root gives√(S −A)(S −B)

AB= sin

γ

2.

(Note that γ is between 0 and 180 degrees, soγ

2is between 0 and 90 degrees, and

thus sinγ

2is nonnegative.)

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588 Solution Guide for Chapter 8

10. Heron’s area formula:

(a) Part (c) of Exercise 8 says

S(S − C) =12AB(1 + cos γ),

and Part (c) of Exercise 9 says

(S −A)(S −B) =12AB(1− cos γ).

Thus, multiplying the earlier results gives

S(S−A)(S−B)(S−C) =12AB(1+cos γ)

12AB(1−cos γ) =

14A2B2(1+cos γ)(1−cos γ).

Now

(1 + cos γ)(1− cos γ) = 1− cos2 γ = sin2 γ.

Hence

S(S −A)(S −B)(S − C) =14A2B2(1 + cos γ)(1− cos γ) =

14A2B2 sin2 γ.

(b) The area of the triangle is12AB sin γ, and by Part (a) this equals

√S(S −A)(S −B)(S − C).

11. Getting area: In the notation of Exercise 10, S =12(7 + 8 + 9) = 12, and by Heron’s

formula the area of the triangle is

√12(12− 7)(12− 8)(12− 9) =

√720 = 26.83.

12. Getting area: In the notation of Exercise 10, S =12(6 + 9 + 13) = 14, and by Heron’s

formula the area of the triangle is

√14(14− 6)(14− 9)(14− 13) =

√560 = 23.66.

13. No such triangle: If there were such a triangle then by the law of sines we would have

12sinα

=4

sin 20◦,

so

sinα = 12× sin 20◦

4= 1.03.

But this is impossible because the sine function is never larger than 1.

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SECTION 8.2 Laws of Sines and Cosines 589

14. No such triangle: If there were such a triangle then by the law of sines we would have

11sinα

=1

sin 70◦,

so

sinα = 11× sin 70◦ = 10.34.

But this is impossible because the sine function is never larger than 1.

15. Two such triangles: By the law of sines we have

25sinα

=22

sin 55◦,

so

sinα = 25× sin 55◦

22.

Using the crossing-graphs method with a horizontal span from 0 to 180 degrees gives

two solutions, α = 68.57 degrees and α = 111.43 degrees.

When α = 68.57 degrees we get γ = 180 − 68.57 − 55 = 56.43 degrees (because the

angle sum of the triangle is 180 degrees). By the law of sines,

C

sin 56.43◦=

22sin 55◦

,

so

C = sin 56.43◦ × 22sin 55◦

= 22.38.

When α = 111.43 degrees we get γ = 180− 111.43− 55 = 13.57 degrees. By the law of

sines,C

sin 13.57◦=

22sin 55◦

,

so

C = sin 13.57◦ × 22sin 55◦

= 6.30.

The two triangles are shown below.

γ=56.43o

A=25 B=22

β=55o α=68.57o

C=22.38

γ=13.57o

A=25

B=22

β=55o α=111.43o

C=6.30

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590 Solution Guide for Chapter 8

8.3 INVERSE TRIGONOMETRIC FUNCTIONS

E-1. Adding up lots of terms: The sum of the first 10 terms of the series is

4((

45− 1

239

)− 1

3

(453− 1

2393

)+ · · · − 1

19

(4

519− 1

23919

)).

The calculator’s displayed value for this is 3.141592654, which is the same as the calcu-

lator’s displayed value for π. Adding up the first 10 terms of the alternating sum of the

reciprocals of the odd integers (times 4) gives

4(

1− 13

+ · · · − 119

).

The calculator’s displayed value for this is 3.041839619, which differs from the correct

value for π (and thus with the above value from Machin’s formula) already in the first

decimal place.

E-2. The error: If we add up the first 15 terms in Machin’s formula we will miss the true

value of π by no more than431

(4

531− 1

23931

).

This is about 1.11× 10−22.

E-3. Getting Machin’s formula:

(a) As suggested, we apply the double-angle formula for the tangent twice:

tan(4x) = tan(2(2x)) =2 tan(2x)

1− tan2(2x)=

2(

2 tan x1−tan2 x

)1−

(2 tan x

1−tan2 x

)2 .

Now to simplify this we multiply both the numerator and the denominator of the

big fraction by (1− tan2 x)2. The result is

tan(4x) =2(2 tan x)(1− tan2 x)

(1− tan2 x)2 − (2 tan x)2=

4 tanx(1− tan2 x)(1− tan2 x)2 − 4 tan2 x

.

(b) We use x = arctan15

in Part (a), so tanx =15

:

tan(

4 arctan15

)=

4(

15

) (1−

(15

)2)(1−

(15

)2)2

− 4(

15

)2 =120119

.

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SECTION 8.3 Inverse Trigonometric Functions 591

(c) As suggested, we apply the sum formula for the tangent:

tan(

4 arctan15− arctan

1239

)=

tan(4 arctan 1

5

)+ tan

(− arctan 1

239

)1− tan

(4 arctan 1

5

)tan

(− arctan 1

239

) .Now in general tan(−x) = − tanx because sine is odd and cosine is even. Thus

tan(− arctan

1239

)= − tan

(arctan

1239

)= − 1

239.

Using this and the result of Part (b) gives

tan(

4 arctan15− arctan

1239

)=

120119 −

1239

1 +(

120119

) (1

239

) = 1.

S-1. We want an angle between −90 and 90 degrees whose sine is√

32

. Now sin 60◦ =√

32

,

so arcsin

(√3

2

)equals 60 degrees or

π

3radians.

S-2. By Part 6 of Key Idea 8.5,

arcsin

(−√

32

)= − arcsin

(√3

2

).

In Exercise S-1 we found that arcsin

(√3

2

)equals 60 degrees or

π

3radians. Hence

arcsin

(−√

32

)equals −60 degrees or −π

3radians.

S-3. We want an angle between 0 and 180 degrees whose cosine is√

32

. Now cos 30◦ =√

32

,

so arccos

(√3

2

)equals 30 degrees or

π

6radian.

S-4. By Part 6 of Key Idea 8.5, in terms of degrees

arccos

(−√

32

)= 180◦ − arccos

(√3

2

).

In Exercise S-3 we found that arccos

(√3

2

)= 30◦. Because 180−30 = 150, arccos

(−√

32

)equals 150 degrees or

6radians.

S-5. We want an angle between 0 and 180 degrees whose cosine is 0. Now cos 90◦ = 0, so

arccos 0 equals 90 degrees orπ

2radians.

S-6. By Part 6 of Key Idea 8.5, in terms of degrees

arccos(−1

2

)= 180◦ − arccos

(12

).

Now cos 60◦ =12, so arccos

(12

)= 60◦. Because 180 − 60 = 120, arccos

(−1

2

)equals

120 degrees or2π

3radians.

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592 Solution Guide for Chapter 8

S-7. By Part 6 of Key Idea 8.5,

arcsin(−1

2

)= − arcsin

(12

).

Now sin 30◦ =12, so arcsin

(−1

2

)equals −30 degrees or −π

6radian.

S-8. We want an angle α between −90 and 90 degrees so that tanα = 0. Now tanα =sinα

cos α,

so we look for α so that sinα = 0. Now sin 0◦ = 0, so arctan 0 equals 0 degrees or 0

radians.

S-9. We want an angle α between−90 and 90 degrees so that tanα =√

3. Now tanα =sinα

cos α,

and as noted√

3 =

√3

212

, so we look for α so that sinα =√

32

and cos α =12

. The angle

α = 60◦ satisfies both of these conditions, so arctan√

3 equals 60 degrees orπ

3radians.

S-10. By Part 3 of Key Idea 8.6,

arctan(−√

3)

= − arctan√

3.

In Exercise S-9 we found that arctan√

3 equals 60 degrees orπ

3radians. Hence arctan

(−√

3)

equals −60 degrees or −π

3radians.

1. Using inverse trig functions: Let α = arctan x, so tanα = x. We want to represent this

situation by a right triangle with an acute angle of α so that tanα = x. Because

tanα =OppositeAdjacent

,

we can accomplish this by assigning the length x to the side opposite α and the length 1

to the side adjacent to α. Then by the Pythagorean theorem the hypotenuse is√

1 + x2,

so

cos α =Adjacent

Hypotenuse=

1√1 + x2

.

Thus cos(arctanx) =1√

1 + x2.

2. Using inverse trig functions: Let α = arcsinx, so sinα = x. We want to represent this

situation by a right triangle with an acute angle of α so that sinα = x. Because

sinα =Opposite

Hypotenuse,

we can accomplish this by assigning the length x to the side opposite α and the length 1

to the hypotenuse. Then by the Pythagorean theorem the side adjacent to α is√

1− x2,

so

cot α =AdjacentOpposite

=√

1− x2

x.

Thus cot(arcsin x) =√

1− x2

x.

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SECTION 8.3 Inverse Trigonometric Functions 593

3. Using inverse trig functions: Let α = arccos x, so cos α = x. We want to represent this

situation by a right triangle with an acute angle of α so that cos α = x. Because

cos α =Adjacent

Hypotenuse,

we can accomplish this by assigning the length x to the side adjacent to α and the length

1 to the hypotenuse. Then by the Pythagorean theorem the side opposite α is√

1− x2,

so

tanα =OppositeAdjacent

=√

1− x2

x.

Thus tan(arccos x) =√

1− x2

x.

4. Using inverse trig functions: Let α = arccos(2x), so cos α = 2x. We want to represent

this situation by a right triangle with an acute angle of α so that cos α = 2x. Because

cos α =Adjacent

Hypotenuse,

we can accomplish this by assigning the length 2x to the side adjacent to α and the

length 1 to the hypotenuse. Then by the Pythagorean theorem the side opposite α is√1− (2x)2 =

√1− 4x2, so

sinα =Opposite

Hypotenuse=√

1− 4x2

1=√

1− 4x2.

Thus sin(arccos(2x)) =√

1− 4x2.

5. A basketball player: In this situation the angle α he must incline his eyes is an acute

angle of a right triangle with opposite side of length 10 − 6 = 4 and adjacent side of

length 25. Thus tanα =425

. Then α = arctan(

425

), and the calculator gives the value

arctan(

425

)= 9.09 degrees. Thus the angle is 9.09 degrees. An alternative approach

is to solve tanα =425

using the crossing-graphs method with a horizontal span of 0 to

90 in terms of degrees.

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594 Solution Guide for Chapter 8

6. A soccer player: In this situation the angle α is an angle of a triangle with opposite side

of length 24 and adjacent sides of length 20 and 40. By the law of cosines,

242 = 402 + 202 − 2× 40× 20× cos α,

so

cos α =402 + 202 − 242

2× 40× 20.

Then

α = arccos(

402 + 202 − 242

2× 40× 20

),

and for this the calculator gives the value α = 27.13 degrees. Thus the angle is 27.13

degrees. An alternative approach is to solve

cos α =402 + 202 − 242

2× 40× 20

using the crossing-graphs method with a horizontal span of 0 to 180 in terms of degrees.

7. Solving trigonometric equations: By inspection we can see that t = 1 is one solution

of sin 2t = sin(t + 1). Another way to solve this is to take the arcsine of both sides:

arcsin(sin 2t) = arcsin(sin(t + 1)). (This is legitimate because the sine function takes

values between −1 and 1.) Because both 2t and t + 1 lie between 0 and 90 degrees,

arcsin(sin 2t) = 2t and arcsin(sin(t+1)) = t+1. Thus the equation simplifies to 2t = t+1.

Solving this gives t = 1.

8. Solving trigonometric equations: By the double angle formula for the sine, sin(2t) =

2 sin t cos t. Thus the equation can be written as sin(2t) = 0.5. Taking the arcsine of both

sides gives arcsin(sin(2t)) = arcsin(0.5). Because t lies between 0 and 45 degrees, 2t

lies between 0 and 90 degrees, so arcsin(sin(2t)) = 2t. Hence the equation simplifies to

2t = arcsin(0.5). Now arcsin(0.5) = 30◦ because sin 30◦ = 0.5. Hence 2t = 30◦, and so

t = 15◦.

9. Solving trigonometric equations: Now sec t =1

cos tand tan t =

sin t

cos t, so the given

equation says1

cos t= 2 × sin t

cos t. Multiplying both sides by cos t gives 1 = 2 sin t or

sin t =12

. Taking the arcsine of both sides gives arcsin(sin t) = arcsin(

12

). Because

t lies between 0 and 90 degrees, arcsin(sin t) = t. Thus the equation simplifies to t =

arcsin(

12

). Now arcsin

(12

)= 30◦ because sin 30◦ =

12

. Hence t = 30◦.

10. Solving inverse trig equations: We take the tangent of both sides: tan(arctanx) =

tan 9◦. Now the equation tan(arctanx) = x is always true, so x = tan 9◦ = 0.16.

11. Solving inverse trig equations: We take the sine of both sides: sin(arcsinx) = sinx.

Now the equation sin(arcsinx) = x is always true, so x = sinx.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 595

8.4 COMPLEX NUMBERS AND DeMOIVRE’S THEOREM

E-1. Which graph is it? Because θ is the counterclockwise angle from the positive horizontal

axis, the equation θ = c says that the angle always equals the constant c. This describes

a ray from the origin that makes an angle c with the positive x-axis.

E-2. A missing part: Think of θ as starting at 0 and decreasing, so we are moving in a clock-

wise direction. This corresponds to a second hand on a clock that runs forward as usual,

but with the second hand changing in length as it turns. Its tip will trace out a spiral, but

to describe the spiral we note that the negative values of θ lead to negative values of r,

and we locate these by reflecting through the origin. The result is the spiral obtained by

reflecting through the origin the spiral shown in Figure 8.55. Below we show the graph

of both spirals (using radians). The part for negative values of θ is the darker curve.

E-3. An interesting graph: Experimenting by graphing r = sin(nθ) for different positive

integral values of n leads to the following conclusions: If n is an odd integer, then the

graph looks like a flower with n petals. If n is an even integer, the flower has 2n petals.

Below we show the graph of r = sin(√

2θ) for θ between 0 and 50 using radians. The

graph has overlapping petals that eventually fill the circle of radius 1.

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596 Solution Guide for Chapter 8

E-4. The strophoid: The equation of the strophoid can be written as r =cos(2θ)cos θ

. Below we

show the graph of the strophoid. (Note that we have made the graph in “dot” mode. In

“connected” mode there is an extraneous vertical line x = −1.)

We now describe, using radians to measure θ, how the graph is made. Think of θ as

starting at 0 and increasing, so we are moving in a counterclockwise direction. As the

angle θ increases towardπ

4, the distance r decreases to 0, because the numerator ap-

proaches cos(2× π

4

)= cos

2

)= 0. Hence the graph passes through the origin at

θ =π

4. As we move past θ =

π

4toward θ =

π

2, the numerator cos(2θ) becomes negative

and the denominator cos θ stays positive but approaches 0. Over this range r is negative

and growing in size without limit. Reflecting across the origin to account for the nega-

tive value of r gives the lower part of the graph on the left that goes down. As we move

past θ =π

2, the numerator cos(2θ) stays negative, and the denominator cos θ decreases

from 0 to become negative. Hence r is positive, and no reflection is necessary. This gives

the upper part of the graph on the left. Continuing on through θ =3π

4to θ = π gives

the rest of the graph, which passes through the origin again and returns to the starting

point.

E-5. Coming up with a polar graph: We apply the law of cosines to the triangle in the figure.

The side opposite the angle θ has length x, and the adjacent sides have length12

and

cos θ. Hence

x2 =(

12

)2

+ cos2 θ − 2× 12× cos θ × cos θ =

14

+ cos2 θ − cos2 θ =14.

Thus x2 =14

, so x =12

. Hence the curve is a circle with center(

12, 0)

and radius12

.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 597

S-1. Calculating with complex numbers:

(a) Now (3 + 7i)− (2 + 4i) = (3− 2) + (7− 4)i = 1 + 3i.

(b) Now

(3 + 2i)(5− 6i) = 3× 5− 3× 6i + 2i× 5− 2i× 6i = 15− 8i− 12i2 = 27− 8i

because i2 = −1.

(c) Now

(6 + i)(6− i) = 6× 6− 6× i + i× 6− i× i = 36− i2 = 37

because i2 = −1. This can also be written as 37 + 0i.

S-2. A complex quotient: Following the suggestion, we multiply top and bottom of4 + 2i

1 + iby 1− i:

4 + 2i

1 + i=

4 + 2i

1 + i× 1− i

1− i=

(4 + 2i)(1− i)(1 + i)(i− i)

=4− 2i− 2i2

1− i2=

6− 2i

2= 3− i.

S-3. Locating complex numbers: The numbers are located in the following figure.

(d) 7i

(c) _2+4i

(a) 6+2i

(e) 8

(b) 3-i

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598 Solution Guide for Chapter 8

S-4. Locating complex numbers: The numbers are located in the following figure.

(b)

3

(a)

3π/4 1

π/4

_π/4

1(d)

2(c)

S-5. Standard representation: By the Euler formula

4eiπ/3 = 4 cosπ

3+ i4 sin

π

3= 2 + 2

√3i = 2 + 3.46i

because cosπ

3=

12

and sinπ

3=√

32

(using radians, of course).

S-6. Standard representation: By the Euler formula

6e−iπ/4 = 6 cos(−π

4

)+ i6 sin

(−π

4

)=

6√2− 6√

2i = 4.24− 4.24i,

using radians, of course.

S-7. Standard representation: By the Euler formula

2e4i = 2 cos 4 + i2 sin 4 = −1.31− 1.51i,

using radians, of course.

S-8. Standard representation: As suggested, we use De Moivre’s theorem with n = −1. This

says

(cos θ + i sin θ)−1 = cos(−θ) + i sin(−θ) = cos θ − i sin θ.

Thus1

reiθ=

1r(eiθ)−1 =

1r(cos θ + i sin θ)−1 =

1r

cos θ − i1r

sin θ.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 599

S-9. Verification: Now

(cos 3 + i sin 3)2 = (cos 3 + i sin 3)(cos 3 + i sin 3) = cos2 3− sin2 3 + 2i cos 3 sin 3

because i2 = −1. The calculator gives the values

cos2 3− sin2 3 = 0.96

and

2 cos 3 sin 3 = −0.28.

Thus

(cos 3 + i sin 3)2 = 0.96− 0.28i.

Also, the calculator gives cos 6 = 0.96 and sin 6 = −0.28. Thus

(cos 3 + i sin 3)2 = cos 6 + i sin 6.

S-10. Verification: Now

(cos 1.5+i sin 1.5)2 = (cos 1.5+i sin 1.5)(cos 1.5+i sin 1.5) = cos2 1.5−sin2 1.5+2i cos 1.5 sin 1.5

because i2 = −1. The calculator gives the values

cos2 1.5− sin2 1.5 = −0.99

and

2 cos 1.5 sin 1.5 = 0.14.

Thus

(cos 1.5 + i sin 1.5)2 = −0.99 + 0.14i.

Also, the calculator gives cos 3 = −0.99 and sin 3 = 0.14. Thus

(cos 1.5 + i sin 1.5)2 = cos 3 + i sin 3.

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600 Solution Guide for Chapter 8

1. Standard representation: By the Euler formula

5ei5π/6 = 5 cos5π

6+ 5i sin

6= −5

√3

2+

52i = −4.33 + 2.5i

because cos5π

6= −

√3

2and sin

6=

12

(using radians, of course).

2. Exponential form: The modulus of4√2

+ i4√2

is

r =

√(4√2

)2

+(

4√2

)2

= 4.

We want an angle θ so that

cos θ =4√2

4=

1√2

and

sin θ =4√2

4=

1√2.

Now θ =π

4satisfies this, so an exponential form is 4eπi/4.

3. Exponential form: The modulus of −1 + i√

3 is

r =

√(−1)2 +

(√3)2

= 2.

We want an angle θ so that

cos θ =−12

= −12

and

sin θ =√

32

.

Now θ =2π

3satisfies this, so an exponential form is 2e2πi/3.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 601

4. Powers: Now cosπ

6=√

32

and sinπ

6=

12

using radians. Thus we have

(√3

2+ i

12

)35

=(cos

π

6+ i sin

π

6

)35

.

By De Moivre’s theorem,(cos

π

6+ i sin

π

6

)35

= cos35π

6+ i sin

35π

6.

Now

cos35π

6+ i sin

35π

6= cos

11π

6+ i sin

11π

6=√

32− i

12.

Hence (√3

2+ i

12

)35

=√

32− i

12.

5. Roots: Now cosπ

6=√

32

and sinπ

6=

12

using radians. Thus we want to find the fourth

roots of cosπ

6+ i sin

π

6= eiπ/6. Here they are:

First 4th root = ei 14

π6 = cos

π

24+ i sin

π

24

Second 4th root = ei 14 ( π

6 +2π) = cos13π

24+ i sin

13π

24

Third 4th root = ei 14 ( π

6 +4π) = cos25π

24+ i sin

25π

24

Fourth 4th root = ei 14 ( π

6 +6π) = cos37π

24+ i sin

37π

24.

6. Powers: The calculator gives the values cos 70◦ = 0.34 and sin 70◦ = 0.94. Thus we

have

(0.34 + 0.94i)103 = (cos 70◦ + i sin 70◦)103 .

By De Moivre’s theorem,

(cos 70◦ + i sin 70◦)103 = cos(103× 70◦) + i sin(103× 70◦) = 0.98 + 0.17i.

Hence

(0.34 + 0.94i)103 = 0.98 + 0.17i.

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602 Solution Guide for Chapter 8

7. Powers: The calculator gives the values cos 0.7 = 0.76 and sin 0.7 = 0.64 using radians.

Thus we have

(0.76 + 0.64i)8 = (cos 0.7 + i sin 0.7)8 .

By De Moivre’s theorem,

(cos 0.7 + i sin 0.7)8 = cos(8× 0.7) + i sin(8× 0.7) = 0.78− 0.63i.

Hence

(0.76 + 0.64i)8 = 0.78− 0.63i.

8. Powers: Now cos(−π

3

)=

12

and sin(−π

3

)= −

√3

2using radians. Thus we have

(12− i

√3

2

)30

=(cos(−π

3

)+ i sin

(−π

3

))30

.

By De Moivre’s theorem,(cos(−π

3

)+ i sin

(−π

3

))30

= cos(30×

(−π

3

))+ i sin

(30×

(−π

3

)).

Now

30×(−π

3

)= −10π,

and

cos (−10π) + i sin (−10π) = 1

because −10π is an integral multiple of 2π. Hence(12− i

√3

2

)30

= 1.

9. Roots: Now cos(−π

3

)=

12

and sin(−π

3

)= −

√3

2using radians. Thus we want to find

the fourth roots of cos(−π

3

)+ i sin

(−π

3

)= e−iπ/3. Here they are:

First 4th root = ei 14 (−π

3 ) = cos(− π

12

)+ i sin

(− π

12

)= cos

π

12− i sin

π

12

Second 4th root = ei 14 (−π

3 +2π) = cos5π

12+ i sin

12

Third 4th root = ei 14 (−π

3 +4π) = cos11π

12+ i sin

11π

12

Fourth 4th root = ei 14 (−π

3 +6π) = cos17π

24+ i sin

17π

24.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 603

10. Roots: The calculator gives the values cos 0.6 = 0.83 and sin 0.6 = 0.56 using radians.

Thus we want to find the sixth roots of cos 0.6 + i sin 0.6 = e0.6i. Here they are:

First 6th root = ei 16 (0.6) = cos 0.1 + i sin 0.1

Second 6th root = ei 16 (0.6+2π) = cos

(0.1 +

π

3

)+ i sin

(0.1 +

π

3

)Third 6th root = ei 1

6 (0.6+4π) = cos(

0.1 +2π

3

)+ i sin

(0.1 +

3

)Fourth 6th root = ei 1

6 (0.6+6π) = cos (0.1 + π) + i sin (0.1 + π)

Fifth 6th root = ei 16 (0.6+8π) = cos

(0.1 +

3

)+ i sin

(0.1 +

3

)Sixth 6th root = ei 1

6 (0.6+10π) = cos(

0.1 +5π

3

)+ i sin

(0.1 +

3

).

11. Roots: The calculator gives the values cos 2 = −0.42 and sin 2 = 0.91 using radians.

Thus we want to find the cube roots of cos 2 + i sin 2 = e2i. Here they are:

First cube root = ei 13 (2) = cos

23

+ i sin23

Second cube root = ei 13 (2+2π) = cos

(23

+2π

3

)+ i sin

(23

+2π

3

)Third cube root = ei 1

3 (2+4π) = cos(

23

+4π

3

)+ i sin

(23

+4π

3

).

12. Roots: The values cos π = −1 and sinπ = 0 (in terms of radians) can be found using

the unit-circle definitions or the calculator. Thus we want to find the cube roots of

cos π + i sinπ = eiπ . Here they are:

First cube root = ei 13 (π) = cos

π

3+ i sin

π

3=

12

+ i

√3

2Second cube root = ei 1

3 (π+2π) = cos π + i sinπ = −1

Third cube root = ei 13 (π+4π) = cos

3+ i sin

3=

12− i

√3

2.

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604 Solution Guide for Chapter 8

13. Roots: The values cos3π

2= 0 and sin

2= −1 (in terms of radians) can be found

using the unit-circle definitions or the calculator. Thus we want to find the fifth roots

of cos3π

2+ i sin

2= ei3π/2. Here they are:

First fifth root = ei 15 ( 3π

2 ) = cos3π

10+ i sin

10

Second fifth root = ei 15 ( 3π

2 +2π) = cos7π

10+ i sin

10

Third fifth root = ei 15 ( 3π

2 +4π) = cos11π

10+ i sin

11π

10

Fourth fifth root = ei 15 ( 3π

2 +6π) = cos15π

10+ i sin

15π

10= −i

Fifth fifth root = ei 15 ( 3π

2 +8π) = cos19π

10+ i sin

19π

10.

14. Roots: The values cosπ

4= sin

π

4=

1√2

(in terms of radians) can be found using special

values of sine and cosine or the calculator. Thus we want to find the fourth roots of

cosπ

4+ i sin

π

4= eiπ/4. Here they are:

First fourth root = ei 14 ( π

4 ) = cosπ

16+ i sin

π

16

Second fourth root = ei 14 ( π

4 +2π) = cos9π

16+ i sin

16

Third fourth root = ei 14 ( π

4 +4π) = cos17π

16+ i sin

17π

16

Fourth fourth root = ei 14 ( π

4 +6π) = cos25π

16+ i sin

25π

16.

15. Exponential form: The modulus of 1 + i tan θ is

r =√

12 + (tan θ)2 =√

1 + tan2 θ.

But

tan2 θ = sec2 θ − 1

by Example 8.1 in Section 8.1, so

r =√

1 + sec2 θ − 1 = sec θ.

Factoring this out gives

1 + i tan θ = sec θ × 1 + i tan θ

sec θ= sec θ(cos θ + i sin θ).

Thus the polar form is (sec θ)eθi.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 605

16. The modulus of a complex number: As suggested, by the Euler formula we have

a + bi = reiθ = r cos θ + ir sin θ.

Because two complex numbers are equal exactly when their real and imaginary parts

are equal, we have a = r cos θ and b = r sin θ. Then

a2 + b2 = (r cos θ)2 + (r sin θ)2 = r2(cos2 θ + sin2 θ) = r2.

Hence√

a2 + b2 = r.

17. The complex conjugate:

(a) In the notation of the exercise, a = 1 and b = 2. Thus the complex conjugate of

1 + 2i is 1− 2i.

(b) In the notation of the exercise, a = 1 and b = −3. Thus the complex conjugate of

1− 3i is 1 + 3i.

(c) In the notation of the exercise, a = 7 and b = 0. Thus the complex conjugate of 7 is

7− 0i = 7.

(d) In the notation of the exercise, a = 0 and b = −3. Thus the complex conjugate of

−3i is 0− (−3)i = 3i.

(e) The standard representation of 3eiπ/7 is 3 cosπ

7+ i3 sin

π

7. Hence the complex

conjugate of 3eiπ/7 is

3 cosπ

7− i3 sin

π

7= 3 cos

(−π

7

)+ i3 sin

(−π

7

)= 3e−iπ/7.

18. More on the modulus of a complex number: If a and b are real, then multiplying a+ ib

by its conjugate a− ib gives

(a + bi)(a− ib) = a2 − iab + iab− i2b2 = a2 + b2,

which is the square of the modulus of a + ib.

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606 Solution Guide for Chapter 8

19. Double-angle formulas: By De Moivre’s theorem,

(cos θ + i sin θ)2 = cos 2θ + i sin 2θ.

Direct multiplication gives

(cos θ + i sin θ)2 = (cos θ + i sin θ)(cos θ + i sin θ) = (cos2 θ − sin2 θ) + i(2 sin θ cos θ).

If we equate the two expressions for (cos θ + i sin θ)2 and recall that two complex num-

bers are equal exactly when their real and imaginary parts are equal, we get the double-

angle formulas:

cos 2θ = cos2 θ − sin2 θ

and

sin 2θ = 2 sin θ cos θ.

20. Triple-angle formulas:

(a) By De Moivre’s theorem,

(cos θ + i sin θ)3 = cos 3θ + i sin 3θ.

Direct multiplication gives

(cos θ + i sin θ)3 = (cos θ + i sin θ)2(cos θ + i sin θ)

= ((cos2 θ − sin2 θ) + i(2 sin θ cos θ))(cos θ + i sin θ)

= (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ).

If we equate the two expressions for (cos θ + i sin θ)3 and recall that two complex

numbers are equal exactly when their real and imaginary parts are equal, we get

cos 3θ = cos3 θ − 3 cos θ sin2 θ

sin 3θ = 3 cos2 θ sin θ − sin3 θ.

(b) Because sin2 θ + cos2 θ = 1, we have sin2 θ = 1− cos2 θ. Thus, by Part (a),

cos 3θ = cos3 θ − 3 cos θ sin2 θ = cos3 θ − 3 cos θ(1− cos2 θ) = 4 cos3 θ − 3 cos θ.

In a similar way, from sin2 θ + cos2 θ = 1 we get cos2 θ = 1− sin2 θ. Hence, by Part

(a),

sin 3θ = 3 cos2 θ sin θ − sin3 θ = 3(1− sin2 θ) sin θ − sin3 θ = 3 sin θ − 4 sin3 θ.

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SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 607

21. Sum formulas: As suggested, we start with the equation

ei(α+β) = eiαeiβ .

Expanding the expression on the left using the Euler formula gives

ei(α+β) = cos(α + β) + i sin(α + β).

In a similar way we find

eiαeiβ = (cos α + i sinα)(cos β + i sinβ).

Direct multiplication in the latter expression gives

(cos α + i sinα)(cos β + i sinβ) = (cos α cos β − sinα sinβ) + i(sinα cos β + cos α sinβ).

Now we equate the expressions we found for both sides of the equation

ei(α+β) = eiαeiβ

and recall that two complex numbers are equal exactly when their real and imaginary

parts are equal. The result is

cos(α + β) = cos α cos β − sinα sinβ

and

sin(α + β) = sin α cos β + cos α sinβ.

These are the sum formulas.

22. Lots of values: As suggested, we start with the observation that −1 = ei(π+2kπ) for

every integer k. This implies that, for every integer k,

ei(π+2kπ)×i = e−(π+2kπ)

is a value of (−1)i. This gives infinitely many values of (−1)i and represents a sequence

of positive real numbers with limiting value 0.

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608 Solution Guide for Chapter 8

Chapter 8 Review Exercises

1. Sum formula: Now

sin 105◦ = sin(45◦+60◦) = sin 45◦ cos 60◦+cos 45◦ sin 60◦ =1√2× 1

2+

1√2×√

32

= 0.97.

2. Area: The area is12× 3× 2× sin 35◦ = 1.72.

3. Identity: We use the half-angle formula

cos2(

t

2

)=

1 + cos t

2

with t = 2x to obtain

cos2 x =1 + cos(2x)

2.

Then

cos4 x = (cos2 x)2 =(

1 + cos(2x)2

)2

=14(1 + 2 cos(2x) + cos2(2x)

).

Now we use the half-angle formula

cos2(

t

2

)=

1 + cos t

2

with t = 4x to obtain

cos2(2x) =1 + cos(4x)

2.

Thus

cos4 x =14

(1 + 2 cos(2x) +

1 + cos(4x)2

)=

38

+12

cos(2x) +18

cos(4x).

This can also be done by starting with the right-hand side and using the double-angle

formula for the cosine.

4. Identity: Now the sum formula says

cos(x + y) = cos x cos y − sinx sin y,

and the difference formula says

cos(x− y) = cos x cos y + sinx sin y.

Adding these gives

cos(x + y) + cos(x− y) = 2 cos x cos y.

Dividing by 2 gives

cos x cos y =12(cos(x + y) + cos(x− y)).

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Chapter 8 Review Exercises 609

5. Sides and angles: We use the law of cosines to find C:

C2 = 32 + 42 − 2× 3× 4 cos 80◦,

so

C =√

32 + 42 − 2× 3× 4 cos 80◦ = 4.56.

Now we use the law of sines to find a (the angle opposite A):

3sin a

=4.56

sin 80◦,

so

sin a = 3× sin 80◦

4.56.

To solve this equation we first note that the angle a must be less than 90 degrees because

A isn’t the longest side of the triangle. Then using the crossing-graphs method or the

arcsine function gives a = 40.38 degrees. Because the angle sum of the triangle is 180

degrees, the angle opposite B is 180− 40.38− 80 = 59.62 degrees.

6. Sides and angles: First we use the law of sines to find b (the angle opposite B):

10sin b

=20

sin 130◦,

so

sin b = 10× sin 130◦

20.

To solve this equation we first note that the angle b must be less than 90 degrees because

the angle opposite A is greater than 90 degrees. Then using the crossing-graphs method

or the arcsine function gives b = 22.52 degrees. Because the angle sum of the triangle

is 180 degrees, the angle opposite C is 180− 130− 22.52 = 27.48 degrees. Now we use

the law of sines to find C:C

sin 27.48◦=

20sin 130◦

,

so

C = sin 27.48◦ × 20sin 130◦

= 12.05.

(We can also find C using the law of cosines.)

7. Slide: The slide forms one side of a triangle for which the opposite angle is 20 degrees,

with the other sides having lengths 50 and 90 (both in feet). By the law of cosines the

length of the slide is√502 + 902 − 2× 50× 90 cos 20◦ = 46.29 feet.

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610 Solution Guide for Chapter 8

8. Pole: The top of the pole and the two anchoring points form a triangle. Because the

angle sum of the triangle is 180 degrees, the angle at the top is 180 − 60 − 70 = 50

degrees. Now we use the law of sines to find the lengths of the wires. If A is the length

(in feet) of the wire opposite the 60-degree angle, then

A

sin 60◦=

7.5sin 50◦

,

so

A = sin 60◦ × 7.5sin 50◦

= 8.48 feet.

If B is the length (in feet) of the wire opposite the 70-degree angle, then

B

sin 70◦=

7.5sin 50◦

,

so

B = sin 70◦ × 7.5sin 50◦

= 9.20 feet.

9. Inverse sine: We want an angle between −90 and 90 degrees whose sine is 0. Now

sin 0◦ = 0, so arcsin 0 equals 0 degrees or 0 radians.

10. Inverse tangent: We want an angle α between −90 and 90 degrees so that tanα = 1.

Now tanα =sinα

cos α, so we look for α so that sinα = cos α. Now sin 45◦ = cos 45◦ =

1√2

,

so arctan 1 equals 45 degrees orπ

4radian.

11. Evaluate: Let α = arccos x, so cos α = x. We want to represent this situation by a right

triangle with an acute angle of α so that cos α = x. Because

cos α =Adjacent

Hypotenuse,

we can accomplish this by assigning the length x to the side adjacent to α and the length

1 to the hypotenuse. Then by the Pythagorean theorem the side opposite α is√

1− x2,

so

cot α =AdjacentOpposite

=x√

1− x2.

Thus cot(arccos x) =x√

1− x2.

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Chapter 8 Review Exercises 611

12. Solve: Let α = arccos x, so x = cos α. Now the arccosine takes values between 0 and

180 degrees, and we are told that x is positive and less than 1, so α is positive and less

than 90 degrees. In terms of α the original equation says

sin(2α) = sin α.

By the double-angle formula for the sine, this says

2 sinα cos α = sinα.

By the restrictions on α we know that sinα is not zero. Hence we can divide both sides

by 2 sinα to get cos α =12

. Thus the solution is x =12

.

13. Calculating with complex numbers:

(a) Now (−2 + 3i)− (5 + i) = (−2− 5) + (3− 1)i = −7 + 2i.

(b) Now

i(5 + 7i) = i× 5 + i× 7i = 5i + 7i2 = −7 + 5i

because i2 = −1.

(c) Now

(1− i)2 = (1− i)(1− i) = 1× 1− 1× i− i× 1 + i× i = 1− 2i + i2 = −2i

because i2 = −1.

14. Standard representation: By the Euler formula

7e−3iπ/2 = 7 cos(−3π

2

)+ 7i sin

(−3π

2

)= 7i

because (by the unit-circle definition or the calculator) cos(−3π

2

)= 0 and sin

(−3π

2

)=

1 (using radians, of course).

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612 Solution Guide for Chapter 8

15. Powers: Because cos(−π

4

)=

1√2

and sin(−π

4

)=

1√2

(using radians), we have

(1√2− i

1√2

)4

=(cos(−π

4

)+ i sin

(−π

4

))4

.

By De Moivre’s theorem,(cos(−π

4

)+ i sin

(−π

4

))4

= cos(4×

(−π

4

))+ i sin

(4×

(−π

4

))= cos (−π) + i sin (−π)

= −1.

Hence (1√2− i

1√2

)4

= −1.

16. Roots: Now cosπ

2= 0 and sin

π

2= 1 using radians. Thus we want to find the square

roots of cosπ

2+ i sin

π

2= eiπ/2. Here they are:

First square root = ei 12 ( π

2 ) = cosπ

4+ i sin

π

4=

1√2

+ i1√2

Second square root = ei 12 ( π

2 +2π) = cos5π

4+ i sin

4= − 1√

2− i

1√2.