## Engage NY Eureka Math Precalculus Module 4 Lesson 3 Answer Key

### Eureka Math Precalculus Module 4 Lesson 3 Example Answer Key

Example 1.
Consider the figures below. The figure on the right is obtained from the figure on the left by rotating by – β about the origin. a. Calculate the length of $$\overline{A B}$$ in the figure on the left. b. Calculate the length of $$\overline{A}^{\prime} {B}^{\prime}$$ in the figure on the right. c. Set AB and A’B’ equal to each other, and solve the equation for cos(α – β). Example 2.
Use the identity tan(θ) = $$\frac{\sin (\theta)}{\cos (\theta)}$$ to show that tan(α + β) = $$\frac{\tan (\alpha) + \tan (\beta)}{1 – \tan (\alpha) \tan (\beta)}$$. ### Eureka Math Precalculus Module 4 Lesson 3 Exercise Answer Key

Exercises 1–2
Exercises 1.
Use the fact that cos( – θ) = cos(θ) to determine a formula for cos(α + β).
cos(α + β) = cos(α – β)
= cos(α)cos( – β) + sin(α)sin( – β)
= cos(α)cos(β) + sin(α)( – sin(β))
= cos(α)cos(β) – sin(α)sin(β)

Exercises 2.
Use the fact that sin(θ) = cos($$\frac{\pi}{2}$$ – θ) to determine a formula for sin(α – β).
sin(α – β) = cos⁡($$\frac{\pi}{2}$$ – (α – β))
= cos⁡(($$\frac{\pi}{2}$$ – α) + β)
= cos($$\frac{\pi}{2}$$ – α) cos⁡(β) – sin($$\frac{\pi}{2}$$ – α) sin⁡(β)
= sin(α)cos(β) – cos(α)sin(β)

Exercises 3–5
Exercises 3.
Verify the identity tan(α – β) = $$\frac{\tan (\alpha) – \tan (\beta)}{1 + \tan (\alpha) \tan (\beta)}$$ for all (α – β)≠$$\frac{\pi}{2}$$ + πn. Exercises 4.
Use the addition and subtraction formulas to evaluate the expressions shown.
a. cos( – $$5\frac{\pi}{12}$$)
cos( – $$\frac{5\pi}{12}$$) = cos($$\frac{\pi}{4}$$ – $$\frac{2\pi}{3}$$)
= cos($$\frac{\pi}{4}$$)cos($$\frac{2\pi}{3}$$) + sin($$\frac{\pi}{4}$$)sin($$\frac{2\pi}{3}$$)
= $$\frac{\sqrt{2}}{2}$$ ($$\frac{ – 1}{2}$$) + $$\frac{\sqrt{2}}{2}$$ ($$\frac{\sqrt{3}}{2}$$)
= $$\frac{\sqrt{6} – \sqrt{2}}{4}$$

b. sin($$\frac{23\pi}{12}$$)
sin($$\frac{23\pi}{12}$$) = sin($$\frac{9\pi}{4}$$ – $$\frac{\pi}{3}$$)
= sin($$\frac{9\pi}{4}$$)cos($$\frac{\pi}{3}$$) – sin($$\frac{\pi}{3}$$)cos($$\frac{9\pi}{4}$$)
= $$\frac{\sqrt{2}}{2}$$ ($$\frac{1}{2}$$) – $$\frac{\sqrt{3}}{2}$$ ($$\frac{\sqrt{2}}{2}$$)
= $$\frac{\sqrt{2} – \sqrt{6}}{4}$$

c. tan($$\frac{5\pi}{12}$$) Exercises 5.
Use the addition and subtraction formulas to verify these identities for all real – number values of θ.
a. sin(π – θ) = sin(θ)
sin(π – θ) = sin(π)cos(θ) – sin(θ)cos(π) = 0(cos(θ)) – sin(θ)( – 1) = sin(θ)

b. cos(π + θ) = – cos(θ)
cos(π + θ) = cos(π)cos(θ) – sin(π)sin(θ) = – 1(cos(θ)) – 0(sin(θ)) = – cos(θ)

### Eureka Math Precalculus Module 4 Lesson 3 Problem Set Answer Key

Question 1.
Use the addition and subtraction formulas to evaluate the given trigonometric expressions.
a. cos($$\frac{\pi}{12}$$)
cos($$\frac{\pi}{12}$$) = cos($$\frac{\pi}{3}$$ – $$\frac{\pi}{4}$$)
= cos($$\frac{\pi}{3}$$)cos($$\frac{\pi}{4}$$) + sin($$\frac{\pi}{3}$$)sin($$\frac{\pi}{4}$$)
= $$\frac{1}{2}$$∙$$\frac{\sqrt{2}}{2}$$ + $$\frac{\sqrt{3}}{2}$$∙$$\frac{\sqrt{2}}{2}$$
= $$\frac{\sqrt{2} + \sqrt{6}}{4}$$

b. sin($$\frac{\pi}{12}$$)
sin($$\frac{\pi}{12}$$) = sin($$\frac{\pi}{3}$$ – $$\frac{\pi}{4}$$)
= sin($$\frac{\pi}{3}$$)cos($$\frac{\pi}{4}$$) – cos($$\frac{\pi}{3}$$)sin($$\frac{\pi}{4}$$)
= $$\frac{\sqrt{3}}{2}$$∙$$\frac{\sqrt{2}}{2}$$ – $$\frac{1}{2}$$∙$$\frac{\sqrt{2}}{2}$$
= $$\frac{\sqrt{6} – \sqrt{2}}{4}$$

c. sin(5$$\frac{\pi}{12}$$)
sin(5$$\frac{\pi}{12}$$) = sin($$\frac{\pi}{6}$$ + $$\frac{\pi}{4}$$)
= sin($$\frac{\pi}{6}$$)cos($$\frac{\pi}{4}$$) + cos($$\frac{\pi}{6}$$)sin($$\frac{\pi}{4}$$)
= $$\frac{1}{2}$$∙$$\frac{\sqrt{2}}{2}$$ + $$\frac{\sqrt{3}}{2}$$∙$$\frac{\sqrt{2}}{2}$$
= $$\frac{\sqrt{2} + \sqrt{6}}{4}$$

d. cos( – $$\frac{\pi}{12}$$)
cos( – $$\frac{\pi}{12}$$) = cos($$\frac{\pi}{4}$$ – $$\frac{\pi}{3}$$)
= cos($$\frac{\pi}{4}$$)cos($$\frac{\pi}{3}$$) + sin($$\frac{\pi}{4}$$)sin($$\frac{\pi}{3}$$)
= $$\frac{\sqrt{2}}{2}$$∙$$\frac{1}{2}$$ + $$\frac{\sqrt{2}}{2}$$∙$$\frac{\sqrt{3}}{2}$$
= $$\frac{\sqrt{2} + \sqrt{6}}{4}$$

e. sin($$\frac{7\pi}{12}$$)
sin($$\frac{7\pi}{12}$$) = sin($$\frac{\pi}{4}$$ + $$\frac{\pi}{3}$$)
= sin($$\frac{\pi}{4}$$)cos($$\frac{\pi}{3}$$) + cos($$\frac{\pi}{4}$$)sin($$\frac{\pi}{3}$$)
= $$\frac{\sqrt{2}}{2}$$∙$$\frac{1}{2}$$ + $$\frac{\sqrt{2}}{2}$$∙$$\frac{\sqrt{3}}{2}$$
= $$\frac{\sqrt{2} + \sqrt{6}}{4}$$

f. cos( – $$7\frac{\pi}{12}$$)
cos( – $$\frac{7\pi}{12}$$) = cos( – $$\frac{\pi}{4}$$ – $$\frac{\pi}{3}$$)
= cos( – $$\frac{\pi}{4}$$)cos($$\frac{\pi}{3}$$) + sin( – $$\frac{\pi}{4}$$)sin($$\frac{\pi}{3}$$)
= $$\frac{\sqrt{2}}{2}$$∙$$\frac{1}{2}$$ – $$\frac{\sqrt{2}}{2}$$∙$$\frac{\sqrt{3}}{2}$$
= $$\frac{\sqrt{2} – \sqrt{6}}{4}$$

g. sin($$\frac{13\pi}{12}$$)
sin($$\frac{13\pi}{12}$$) = sin($$\frac{3\pi}{4}$$ + $$\frac{\pi}{3}$$)
= sin($$\frac{3\pi}{4}$$)cos($$\frac{\pi}{3}$$) + cos($$\frac{3\pi}{4}$$)sin($$\frac{\pi}{3}$$)
= $$\frac{\sqrt{2}}{2}$$∙$$\frac{1}{2}$$ – $$\frac{\sqrt{2}}{2}$$∙$$\frac{\sqrt{3}}{2}$$
= $$\frac{\sqrt{2} – \sqrt{6}}{4}$$

h. cos( – $$\frac{13\pi}{12}$$)
cos( – $$\frac{13\pi}{12}$$) = cos( – $$\frac{\pi}{4}$$ – $$\frac{\pi}{3}$$)
= cos( – $$\frac{\pi}{4}$$)cos($$\frac{\pi}{3}$$) + sin( – $$\frac{\pi}{4}$$)sin($$\frac{\pi}{3}$$)
= $$\frac{\sqrt{2}}{2}$$∙$$\frac{1}{2}$$ – $$\frac{\sqrt{2}}{2}$$∙$$\frac{\sqrt{3}}{2}$$
= $$\frac{\sqrt{2} – \sqrt{6}}{4}$$

i. sin($$\frac{\pi}{12}$$)cos($$\frac{\pi}{12}$$) + cos($$\frac{\pi}{12}$$)sin($$\frac{\pi}{12}$$)
sin($$\frac{\pi}{12}$$)cos($$\frac{\pi}{12}$$) + cos($$\frac{\pi}{12}$$)sin($$\frac{\pi}{12}$$) = sin($$\frac{\pi}{6}$$)
= $$\frac{1}{2}$$

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

k. sin$$\frac{\pi}{8}$$)cos$$\frac{\pi}{8}$$) + cos$$\frac{\pi}{8}$$)sin$$\frac{\pi}{8}$$)
sin$$\frac{\pi}{8}$$)cos$$\frac{\pi}{8}$$) + cos$$\frac{\pi}{8}$$)sin$$\frac{\pi}{8}$$) = sin($$\frac{\pi}{4}$$)
= $$\frac{\sqrt{2}}{2}$$

l. cos$$\frac{\pi}{8}$$)cos$$\frac{\pi}{8}$$) – sin$$\frac{\pi}{8}$$)sin$$\frac{\pi}{8}$$)
cos$$\frac{\pi}{8}$$)cos$$\frac{\pi}{8}$$) – sin$$\frac{\pi}{8}$$)sin$$\frac{\pi}{8}$$) = cos($$\frac{\pi}{4}$$)
= $$\frac{\sqrt{2}}{2}$$

m. cos($$\frac{\pi}{4}$$)cos($$\frac{\pi}{12}$$) + sin($$\frac{\pi}{4}$$)sin($$\frac{\pi}{12}$$)
cos($$\frac{\pi}{4}$$)cos($$\frac{\pi}{12}$$) + sin($$\frac{\pi}{4}$$)sin($$\frac{\pi}{12}$$) = cos($$\frac{\pi}{6}$$)
= $$\frac{\sqrt{3}}{2}$$

n. sin(π/3)cos($$\frac{\pi}{12}$$) – cos(π/3)sin($$\frac{\pi}{12}$$)
sin(π/3)cos($$\frac{\pi}{12}$$) – cos(π/3)sin($$\frac{\pi}{12}$$) = sin($$\frac{\pi}{4}$$)
= $$\frac{\sqrt{2}}{2}$$

Question 2.
Figure 2 is obtained from Figure 1 by rotating the angle by α about the origin.
Use the method shown in Example 1 to show that cos(α + β) = cos(α)cos(β) – sin(α)sin(β).
Figure 1 Figure 2 AB = $$\sqrt{(\cos (\beta) – \cos (\alpha))^{2} + (\sin (\beta) – \sin (\alpha))^{2}}$$
A’B’ = $$\sqrt{(\cos (\alpha + \beta) – ( – 1))^{2} + (\sin (\alpha + \beta) – 0)^{2}}$$
AB = A’B’
$$\left(\sqrt{(\cos (\beta) – \cos (\alpha))^{2} + (\sin (\beta) – \sin (\alpha))^{2}}\right)^{2}$$ = $$\left(\sqrt{(\cos (\alpha + \beta) – ( – 1))^{2} + (\sin (\alpha + \beta) – 0)^{2}}\right)^{2}$$
(cos(β) – cos(α))2 + (sin(β) – sin(α))2 = (cos(α + β) – ( – 1))2 + (sin(α + β) – 0)2
cos2 (β) – 2 cos(β)cos(α) + cos2 (α) + sin2 (β) – 2 sin(β)sin(α) + sin2 (α) = cos2 (α + β) + 2 cos(α + b) + 1 + sin2 (α + β)
2 – 2cos(β)cos(α) – 2sin(β)sin(α) = 2 + 2 cos(α + β)
– cos(β)cos(α) – sin(β)sin(α) = cos(α + β)
cos(α)cos(β) – sin(α)sin(β) = cos(α + β)

Question 3.
Use the sum formula for sine to show that sin(α – β) = sin(α)cos(β) – cos(α)sin(β).
sin(α – β) = sin(α + ( – β)) = sin(α)cos( – β) + cos(α)sin( – β) = sin(α)cos(β) – cos(α)sin(β)

Question 4.
Evaluate tan(α + β) = $$\frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)}$$ to show tan(α + β) = $$\frac{\tan (\alpha) + \tan (\beta)}{1 – \tan (\alpha) \tan (\beta)}$$. Use the resulting formula to show that tan(2α) = $$\frac{2 \tan (\alpha)}{1 – \tan ^{2}(\alpha)}$$. Question 5.
Show tan(α – β) = $$\frac{\tan (\alpha) – \tan (\beta)}{1 + \tan (\alpha) \tan (\beta)}$$. Question 6.
Find the exact value of the following by using addition and subtraction formulas.
a. tan($$\frac{\pi}{12}$$) b. tan( – $$\frac{\pi}{12}$$) c. tan($$\frac{7\pi}{12}$$) d. tan( – $$\frac{13\pi}{12}$$) e. $$\frac{\tan \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{12}\right)}{1 – \tan \left(\frac{\pi}{4}\right) \tan \left(\frac{\pi}{12}\right)}$$ f. $$\frac{\tan \left(\frac{\pi}{3}\right) – \tan \left(\frac{\pi}{12}\right)}{1 + \tan \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{12}\right)}$$ g. $$\frac{\tan \left(\frac{\pi}{12}\right) + \tan \left(\frac{\pi}{12}\right)}{1 – \tan \left(\frac{\pi}{12}\right) \tan \left(\frac{\pi}{12}\right)}$$ ### Eureka Math Precalculus Module 4 Lesson 3 Exit Ticket Answer Key

Question 1.
Prove that sin⁡(α + β) = sin⁡(α) cos⁡(β) + sin⁡(β)cos(α).
sin(α + β) = sin(α – ( – β))
= sin(α)cos( – β) – sin( – β)cos(α)
= sin(α)cos(β) – ( – sin(β))(cos(α))
= sin(α)cos(β) + sin(β)cos(α)

Question 2.
Use the addition and subtraction formulas to evaluate the given trigonometric expressions.
a. sin($$\frac{\pi}{12}$$) b. tan($$\frac{13\pi}{12}$$) 