## Engage NY Eureka Math Precalculus Module 4 Lesson 4 Answer Key

### Eureka Math Precalculus Module 4 Lesson 4 Exercise Answer Key

Exercises
Exercise 1.
Derive formulas for the following:
a. sin(2θ)
sin(2θ) = sin(θ + θ)
= sin(θ)cos(θ) + cos(θ)sin(θ)
= 2sin(θ)cos(θ)

b. cos(2θ)
cos(2θ) = cos(θ + θ)
= cos(θ)cos(θ) – sin(θ)sin(θ)
= cos2 (θ) – sin2 (θ)

Exercise 2.
Use the double – angle formulas for sine and cosine to verify these identities:
a. tan(2θ) = $$\frac{2 \tan (\theta)}{1 – \tan ^{2}(\theta)}$$

b. sin2 (θ) = (1 – cos(2θ))/2
$$\frac{1 – \cos (2 \theta)}{2}$$ = $$\frac{1 – \left(1 – 2 \sin ^{2}(\theta)\right)}{2}$$ = $$\frac{2 \sin ^{2}(\theta)}{2}$$ = sin2 (θ)

c. sin(3θ) = – 4sin3 (θ) + 3sin(θ)
sin(3θ) = sin(2θ + θ)
= sin(2θ)cos(θ) + cos(2θ)sin(θ)
= 2sin(θ)(cos(θ))(cos(θ)) + (1 – 2sin2 (θ))(sin(θ))
= 2sin(θ) cos2 (θ) + sin(θ) – 2sin3 (θ)
= 2sin(θ)(1 – sin2 (θ)) + sin(θ) – 2sin3 (θ)
= 2sin(θ) – 2sin3 (θ) + sin(θ) – 2sin3 (θ)
= – 4sin3 (θ) + 3sin(θ)

Exercise 3.
Suppose that the position of a rider on the unit circle carousel is (0.8, – 0.6) for a rotation θ. What is the position of the rider after rotation by 2θ?
x = cos(2θ) = cos2(θ) – sin2 (θ) = 0.82 – ( – 0.6)2 = 0.64 – 0.36 = 0.28
y = sin(2θ) = 2sin(θ)cos(θ) = 2( – 0.6)(0.8) = – 0.96
The rider’s position is (0.28, – 0.96).

Exercise 4.
Use the double – angle formula for cosine to establish the identity cos($$\frac{\theta}{2}$$) = ±$$\sqrt{\frac{\cos (\theta) + 1}{2}}$$
Since θ = 2($$\frac{\theta}{2}$$), the double – angle formula gives cos(2($$\frac{\theta}{2}$$)) = 2cos2 ($$\frac{\theta}{2}$$) – 1. Then we have
cos(θ) = 2cos2 ($$\frac{\theta}{2}$$) – 1
1 + cos(θ) = 2cos2 ($$\frac{\theta}{2}$$)
$$\frac{1 + \cos (\theta)}{2}$$ = cos2 ($$\frac{\theta}{2}$$)
cos($$\frac{\theta}{2}$$) = ±$$\sqrt{\frac{\cos (\theta) + 1}{2}}$$

Exercise 5.
Use the double – angle formulas to verify these identities:
a. sin($$\frac{\theta}{2}$$) = ±$$\sqrt{\frac{1 – \cos (\theta)}{2}}$$
Since θ = 2($$\frac{\theta}{2}$$), the double – ange formulas give cos(2($$\frac{\theta}{2}$$)) = 1 – 2sin2 ($$\frac{\theta}{2}$$).
Then we have
cos(θ) = 1 – 2sin2 ($$\frac{\theta}{2}$$)
1 – cos(θ) = 2sin2 ($$\frac{\theta}{2}$$)
$$\frac{1 – \cos (\theta)}{2}$$ = sin2 ($$\frac{\theta}{2}$$)
sin($$\frac{\theta}{2}$$) = ±$$\sqrt{\frac{1 – \cos (\theta)}{2}}$$

b. tan($$\frac{\theta}{2}$$) = ±
tan($$\frac{\theta}{2}$$) = $$\frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)}$$ = $$\frac{\pm \sqrt{\frac{1 – \cos (\theta)}{2}}}{\pm \sqrt{\frac{\cos (\theta) + 1}{2}}}$$ = ±$$\sqrt{\frac{1 – \cos (\theta)}{1 + \cos (\theta)}}$$

Exercise 6.
The position of a rider on the unit circle carousel is (0.8, – 0.6) after a rotation by θ where 0 ≤ θ < 2π. What is the position of the rider after rotation by $$\frac{\theta}{2}$$?
Given that cos(θ) is positive and sin(θ) is negative, the rider is located in Quadrant IV after rotation by θ, so
$$\frac{3\pi}{2}$$ < θ < 2π. This means that $$\frac{3\pi}{4}$$ < $$\frac{\theta}{2}$$ < π, which is in Quadrant II, so cos($$\frac{\theta}{2}$$) is negative and sin($$\frac{\theta}{2}$$) is positive.
x($$\frac{\theta}{2}$$) = cos($$\frac{\theta}{2}$$) = ±$$\sqrt{\frac{\cos (\theta) + 1}{2}}$$ = – $$\sqrt{\frac{0.8 + 1}{2}}$$ ≈ – 0.95
y($$\frac{\theta}{2}$$) = sin($$\frac{\theta}{2}$$) = ±$$\sqrt{\frac{1 – \cos (\theta)}{2}}$$ = $$\sqrt{\frac{1 – 0.8}{2}}$$ ≈ 0.32
The rider’s position is approximately ( – 0.95,0.32) after rotation by $$\frac{\theta}{2}$$.

Exercise 7.
Evaluate the following trigonometric expressions.
a. sin($$\frac{3\pi}{8}$$)

b. tan($$\frac{\pi}{24}$$)

### Eureka Math Precalculus Module 4 Lesson 4 Problem Set Answer Key

Question 1.
Evaluate the following trigonometric expressions.
a. 2 sin($$\frac{\pi}{8}$$)cos($$\frac{\pi}{8}$$)
sin($$\frac{\pi}{4}$$) = $$\frac{\sqrt{2}}{2}$$

b. $$\frac{1}{2}$$ sin($$\frac{\pi}{12}$$)cos($$\frac{\pi}{12}$$)
$$\frac{1}{4}$$ (sin($$\frac{\pi}{6}$$)) = $$\frac{1}{8}$$

c. 4 sin( – $$\frac{5\pi}{12}$$)cos( – $$\frac{5\pi}{12}$$)
2 sin($$\frac{ – 5\pi}{6}$$) = – 2sin($$\frac{\pi}{6}$$) = – 2($$\frac{1}{2}$$) = – 1

d.cos2 ($$\frac{3\pi}{8}$$) – sin2 ($$\frac{3\pi}{8}$$)
cos($$\frac{3\pi}{4}$$) = – cos($$\frac{\pi}{4}$$) = – $$\frac{\sqrt{2}}{2}$$

e. 2 cos2 ($$\frac{\pi}{12}$$) – 1
cos($$\frac{\pi}{6}$$) = $$\frac{\sqrt{3}}{2}$$

f. 1 – 2sin2 ( – $$\frac{\pi}{8}$$)
cos( – $$\frac{\pi}{4}$$) = $$\frac{\sqrt{2}}{2}$$

g. cos2 ( – $$\frac{11\pi}{12}$$) – 2
$$\frac{1}{2}$$ (2 cos2 ( – $$\frac{11\pi}{12}$$) – 1) – $$\frac{3}{2}$$ = $$\frac{1}{2}$$ (cos($$\frac{ – 11\pi}{12}$$)) – $$\frac{3}{2}$$ = $$\frac{1}{2}$$ (cos($$\frac{\pi}{6}$$)) – $$\frac{3}{2}$$ = $$\frac{1}{2}$$ ($$\frac{\sqrt{3}}{2}$$) – $$\frac{3}{2}$$ = $$\frac{\sqrt{3}}{4}$$ – $$\frac{3}{2}$$

h. $$\frac{2 \tan \left(\frac{\pi}{8}\right)}{1 – \tan ^{2}\left(\frac{\pi}{8}\right)}$$
tan($$\frac{\pi}{4}$$) = 1

i. $$\frac{2 \tan \left( – \frac{5 \pi}{12}\right)}{1 – \tan ^{2}\left( – \frac{5 \pi}{12}\right)}$$
tan( – $$\frac{5\pi}{6}$$) = tan($$\frac{\pi}{6}$$) = $$\frac{1}{\sqrt{3}}$$ = $$\frac{\sqrt{2}}{3}$$

j. cos2 ($$\frac{\pi}{8}$$)
cos2 ($$\frac{\pi}{8}$$) = $$\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}$$ = $$\frac{1 + \frac{\sqrt{2}}{2}}{2}$$ = $$\frac{1}{2}$$ + $$\frac{\sqrt{2}}{4}$$

k. cos($$\frac{\pi}{8}$$)
Rotation by θ = $$\frac{\pi}{8}$$ terminates in Quadrant I; therefore, cos($$\frac{\pi}{8}$$) has a positive value.
cos($$\frac{\pi}{8}$$) = $$\sqrt{\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}}$$ = $$\sqrt{\frac{1}{2} + \frac{\sqrt{2}}{4}}$$ = $$\frac{\sqrt{2 + \sqrt{2}}}{2}$$

l. cos( – $$\frac{9\pi}{8}$$)
Rotation by θ = – $$\frac{9\pi}{8}$$ terminates in Quadrant II; therefore, cos( – $$\frac{9\pi}{8}$$) has a negative value.

m. sin2($$\frac{\pi}{12}$$)
sin2($$\frac{\pi}{12}$$) = $$\frac{1 – \cos \left(\frac{\pi}{6}\right)}{2}$$ = $$\frac{1 – \frac{\sqrt{3}}{2}}{2}$$ = $$\frac{1}{2}$$ – $$\frac{\sqrt{3}}{4}$$

n. sin($$\frac{\pi}{12}$$)
Rotation by θ = $$\frac{\pi}{12}$$ terminates in Quadrant I; therefore, sin($$\frac{\pi}{12}$$) has a positive value.
sin($$\frac{\pi}{12}$$) = $$\sqrt{\frac{1 – \cos \left(\frac{\pi}{6}\right)}{2}}$$ = $$\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}$$ = $$\frac{\sqrt{2 + \sqrt{3}}}{2}$$

o. sin( – 5$$\frac{\pi}{12}$$)
Rotation by θ = – 5$$\frac{\pi}{12}$$ terminates in Quadrant IV; therefore, sin( – 5$$\frac{\pi}{12}$$) has a negative value.

p. tan($$\frac{\pi}{8}$$)
Rotation by θ = $$\frac{\pi}{8}$$ terminates in Quadrant I; therefore, tan($$\frac{\pi}{8}$$) has a positive value.

q. tan($$\frac{\pi}{12}$$)

r. tan( – $$\frac{3\pi}{8}$$)
Rotation by θ = – $$\frac{3\pi}{8}$$ terminates in Quadrant IV; therefore, tan( – $$\frac{3\pi}{8}$$) has a negative value.

Question 2.
Show that sin(3x) = 3sin(x)cos2 (x) – sin3 (x). (Hint: Use sin(2x) = 2sin(x)cos(x) and the sine sum formula.)
sin(3x) = sin(x + (2x))
= sin(x)cos(2x) + cos(x)sin(2x)
= sin(x)[cos2 (x) – sin2 (x)] + cos(x)[2sin(x)cos(x)]
= sin(x)cos2 (x) – sin3 (x) + 2sin(x)cos2 (x)
= 3sin(x)cos2 (x) – sin3 (x)

Question 3.
Show that cos(3x) = cos3 (x) – 3sin2 (x)cos(x). (Hint: Use cos(2x) = cos2 (x) – sin2 (x) and the cosine sum formula.)
cos(3x) = cos(x + (2x))
= cos(x)cos(2x) – sin(x)sin(2x)
= cos(x)[cos2 (x) – sin2 (x)] – sin(x)[2sin(x)cos(x)]
= cos3 (x) – cos(x)sin2 (x) – 2cos(x)sin2 (x)
= cos3 (x) – 3cos(x)sin2 (x)

Question 4.
Use cos(2x) = cos2 (x) – sin2 (x) to establish the following formulas.
a. cos2 (x) = $$\frac{1 + \cos (2 x)}{2}$$
cos(2x) = cos2 (x) – sin2 (x)
= cos2 (x) – (1 – cos2 (x))
= 2cos2 (x) – 1
Therefore, cos2 (x) = $$\frac{1 + \cos (2 x)}{2}$$

b. sin2 (x) = $$\frac{1 – \cos (2 x)}{2}$$
cos(2x) = cos2 (x) – sin2 (x)
= (1 – sin2 (x)) – sin2 (x)
= 1 – 2sin2 (x)
Therefore, sin2 (x) = $$\frac{1 – \cos (2 x)}{2}$$

Question 5.
Jamia says that because sine is an odd function, sin($$\frac{\theta}{2}$$) is always negative if θ is negative. That is, she says that for negative values of sin($$\frac{\theta}{2}$$) = – $$\sqrt{\frac{1 – \cos (\theta)}{2}}$$. Is she correct? Explain how you know.
Jamia is not correct. Consider θ = – $$\frac{7\pi}{3}$$. In this case, $$\frac{\theta}{2}$$ = – $$\frac{7\pi}{6}$$, and rotation by – $$\frac{7\pi}{6}$$ terminates in Quadrant II. Thus, sin( – $$\frac{7\pi}{6}$$) is positive.

Question 6.
Ginger says that the only way to calculate sin($$\frac{\pi}{12}$$) is using the difference formula for sine since $$\frac{\pi}{12}$$ = $$\frac{\pi}{3}$$ – $$\frac{\pi}{4}$$. Fred says that there is another way to calculate sin($$\frac{\pi}{12}$$). Who is correct and why?
Fred is correct. We can use the half – angle formula with θ = $$\frac{\pi}{6}$$ to calculate sin($$\frac{\pi}{12}$$).

Question 7.
Henry says that by repeatedly applying the half – angle formula for sine we can create a formula for sin($$\frac{\theta}{n}$$) for any positive integer n. Is he correct? Explain how you know.
Henry is not correct. Repeating this process will only give us formulas for sin($$\frac{\theta}{2^{k}}$$) for positive integers k. There is no way to derive a formula for quantities such as sin($$\frac{\theta}{5}$$) using this method.

### Eureka Math Precalculus Module 4 Lesson 4 Exit Ticket Answer Key

Question 1.
Show that cos⁡(3θ) = 4cos3 (θ) – 3cos⁡(θ).
Evaluate cos($$\frac{7\pi}{12}$$) using the half – angle formula, and then verify your solution using a different formula.