## Engage NY Eureka Math Precalculus Module 4 Lesson 6 Answer Key

### Eureka Math Precalculus Module 4 Lesson 6 Exercise Answer Key

Opening Exercise

When you hear a musical note played on an instrument, the tones are caused by vibrations of the instrument. The vibrations can be represented graphically as a sinusoid. The amplitude is a measure of the loudness of the note, and the frequency is a measure of the pitch of the note. Recall that the frequency of a sinusoidal function is the reciprocal of its period. Louder notes have greater amplitude, and higher pitched notes have larger frequencies.

a. State the amplitude, period, and frequency of each sinusoidal function graphed below.

Answer:

Function f has amplitude 10, period 5, and frequency \(\frac{1}{5}\).

Function g has amplitude 5, period 10, and frequency \(\frac{1}{10}\).

Function h has amplitude 20, period 8, and frequency \(\frac{1}{8}\).

Function k has amplitude 15, period 2, and frequency \(\frac{1}{2}\).

b. Order the graphs from quietest note to loudest note.

Answer:

The quietest notes have lower amplitudes. The order from quietest note to loudest note would be g, f, k, and h.

c. Order the graphs from lowest pitch note to highest pitch note.

Answer:

Lower frequencies are lower pitch. The order from lowest pitch note to highest pitch note would be g, h, f, and k.

Exercises 1–7

When two musical notes are played simultaneously, wave interference occurs. Wave interference is also responsible for the actual sound of the notes that you hear.

Exercise 1.

The graphs of two functions, f and g, are shown below.

a. Model wave interference by picking several points on the graphs of f and g and then using those points to create a graph of h(x) = f(x) + g(x).

Answer:

b. What is a formula for h? Explain how you got your answer.

Answer:

A formula is h(x) = 5sin(x). Since f(x) = sin(x) and g(x) = 4sin(x), you can simply add the two expressions together.

Exercise 2.

The graphs of f and g are shown below.

a. Model wave interference by picking several points on the graphs of f and g and then using those points to create a graph of h(x) = f(x) + g(x).

Answer:

b. What is an approximate formula for h? Explain how you got your answer.

Answer:

From the graph, the amplitude appears to be 5, the period appears to be 2π, and the graph looks like it has been shifted approximately \(\frac{\pi}{4}\) units to the right. So h(x)≈5 cos(x – \(\frac{\pi}{4}\)).

Exercise 3.

Let f(x) = sin(x) and g(x) = cos(x + \(\frac{\pi}{2}\)).

a. Predict what the graph of the wave interference function h(x) = f(x) + g(x) would look like in this situation.

Answer:

The graphs are reflections of one another across the horizontal axis. Therefore, the wave interference function would be h(x) = 0 because the two functions would cancel each other out for every value of x.

b. Use an appropriate identity to confirm your prediction.

Answer:

sin(x) + cos(x + \(\frac{\pi}{2}\)) = sin(x) + cos(x) cos(\(\frac{\pi}{2}\)) – sin(x) sin(\(\frac{\pi}{2}\))

= sin(x) + cos(x)∙0 – sin(x)∙1

= sin(x) – sin(x)

= 0

Exercise 4.

Show that in general, the function h(x) = a cos(bx – c) can be rewritten as the sum of a sine and cosine function with equal periods and different amplitudes.

Answer:

f(x) = a cos(bx – c)

= a(cos(bx) cos(c) + sin(bx) sin(c) )

= a cos(c) cos(bx) + a sin(c) sin(bx)

The two functions would be f(x) = a cos(c) cos(bx) with amplitude |a cos(c) | and period |\(\frac{2\pi}{b}\)| and

g(x) = a sin(c) sin(bx) with amplitude |a sin(c) | and period |\(\frac{2\pi}{b}\)|. The only time the amplitudes would be equal is when c = \(\frac{\pi}{4}\) + πn for integers n.

Exercise 5.

Find an exact formula for h(x) = 12sin(x) + 5cos(x) in the form h(x) = a cos(x – c). Graph f(x) = 12sin(x), g(x) = 5cos(x), and h(x) = 12sin(x) + 5cos(x) together on the same axes.

Answer:

12sin(x) + 5cos(x) = a cos(x – c)

= a(cos(x)cos(c) + sin(x)sin(c))

= a cos(c)cos(x) + a sin(c)sin(x)

Then 12sin(x) = a sin(c) sin(x) and 5cos(x) = a cos(c)cos(x), so we have cos(c) = \(\frac{5}{a}\) and sin(c) = \(\frac{12}{a}\).

Since sin^{2} (c) + cos^{2} (c) = 1, we have \(\frac{144}{a^{2}} + \frac{25}{a^{2}}\) = 1. Thus, 169 = a^{2} and a = 13. Then, tan(c) = \(\frac{12}{5}\), so

c = arctan(\(\frac{12}{5}\))≈1.176. Thus, h(x) = 13cos(x – 1.176).

Exercise 6.

Find an exact formula for h(x) = 2sin(x) – 3cos(x) in the form h(x) = a cos(x – c). Graph f(x) = 2sin(x), g(x) = – 3cos(x), and h(x) = 2sin(x) – 3cos(x) together on the same axes.

Answer:

2sin(x) – 3cos(x) = a cos(x – c)

= a(cos(x)cos(c) + sin(x)sin(c))

= a cos(c)cos(x) + a sin(c)sin(x)

Then, 2sin(x) = a sin(c) sin(x) and – 3cos(x) = a cos(c)cos(x), so we have cos(c) = \(\frac{ – 3}{a}\) and sin(c) = \(\frac{2}{a}\). Since sin^{2} (c) + cos^{2} (c) = 1, we have \(\frac{9}{a^{2}} + \frac{4}{a^{2}}\) = 1. Thus, 13 = a^{2} and a = ±\(\sqrt{13}\). Then, tan(c) = \(\frac{2}{ – 3}\), so c = arctan(\(\frac{2}{ – 3}\))≈ – 0.588. Because h(0) = – 3, and cos( – 0.588)>0, we know that a<0, and thus,

a = – \(\sqrt{13}\). Thus, h(x) = – \(\sqrt{13}\) cos(x + 0.588).

Exercise 7.

Can you find an exact formula for h(x) = 2sin(2x) + 4sin(x) in the form h(x) = a sin(x – c)? If not, why not? Graph f(x) = 2sin(2x), g(x) = 4sin(x), and h(x) = 2sin(2x) + 4sin(x) together on the same axes.

Answer:

Although it is periodic with period 2π, the graph of h is not the graph of a function y = a sin(x – c). Thus, we cannot write the function h in the specified form.

### Eureka Math Precalculus Module 4 Lesson 6 Problem Set Answer Key

Question 1.

Rewrite the sum of the following functions in the form f(x) + g(x) = c cos(x + k). Graph y = f(x), y = g(x), and y = f(x) + g(x) on the same set of axes.

a. f(x) = 4 sin(x); g(x) = 3 cos(x)

Answer:

c = \(\sqrt{(4)^{2} + (3)^{2}}\) = 5,a = arctan(\(\frac{4}{3}\)) ≈ 0.927

f(x) + g(x) = 5 cos(x – 0.927)

b. f(x) = – 6 sin(x); g(x) = 8 cos(x)

Answer:

c = \(\sqrt{( – 6)^{2} + (8)^{2}}\) = 10,a = arctan( – \(\frac{6}{8}\))≈ – 0.644

f(x) + g(x) = 10 cos(x + 0.644)

c. f(x) = \(\sqrt{3}\) sin(x); g(x) = 3 cos(x)

Answer:

c = \(\sqrt{(\sqrt{3})^{2} + (3)^{2}}\) = 2\(\sqrt{3}\),a = arctan \(\left(\frac{\sqrt{3}}{3}\right)\) ≈ 0.524

f(x) + g(x) = 2\(\sqrt{3}\) cos(x – 0.524)

d. f(x) = \(\sqrt{2}\) sin(x); g(x) = \(\sqrt{7}\) cos(x)

Answer:

c = \(\sqrt{(\sqrt{2})^{2} + (\sqrt{7})^{2}}\) = 3, a = arctan(\(\frac{\sqrt{2}}{\sqrt{7}}\)) ≈ 0.491

f(x) + g(x) = 3 cos(x – 0.491)

e. f(x) = 3 sin(x); g(x) = – 2 cos(x)

Answer:

c = \(\sqrt{( – 2)^{2} + (3)^{2}}\) = ±\(\sqrt{13}\),a = arctan(\(\frac{3}{ – 2}\)) ≈ – 0.983

f(x) + g(x) = – \(\sqrt{13}\) cos(x + 0.983)

Question 2.

Find a sinusoidal function f(x) = a sin(bx + c) + d that fits each of the following graphs.

a.

Answer:

f(x) = 3 sin(2x)

b.

Answer:

f(x) = 2 sin(x) + 1

c.

Answer:

f(x) = 3 sin(\(\frac{x}{2}\)) – 1

d.

Answer:

f(x) = – 3 sin(\(\frac{x}{2}\)) – 1

Question 3.

Two functions f and g are graphed below. Sketch the graph of the sum f + g.

a.

Answer:

b.

Answer:

c.

Answer:

d.

Answer:

e.

Answer:

### Eureka Math Precalculus Module 4 Lesson 6 Exit Ticket Answer Key

Question 1.

Use appropriate identities to rewrite the equation shown below in the form h(x) = a cos(x – c).

Answer:

h(x) = 6 sin(x) + 8 cos(x)

6 sin(x) + 8 cos(x) = a cos(c)cos(x) + a sin(c)sin(x)

Then, we have 6 = a sin(c) and 8 = a cos(c), so sin(c) = \(\frac{6}{a}\) and cos(c) = \(\frac{8}{a}\). Since sin^{2} (c) + cos^{2} (c) = 1, we have \(\frac{36}{a^{2}} + \frac{64}{a^{2}}\) = 1, so a = 10. It follows that tan(c) = \(\frac{6}{8}\), so c = arctan(\(\frac{6}{8}\)) ≈ 0.644. Thus,

h(x) = 10 cos(x – 0.644).

Question 2.

Rewrite h(x) = \(\sqrt{3}\) sin(x) + cos(x) in the form h(x) = a cos(x – c).

Answer:

c = \(\sqrt{13}\) = 2

a = tan^{– 1} (\(\sqrt{3}\)) = \(\frac{\pi}{3}\)

h(x) = 2 cos(x – \(\frac{\pi}{3}\))