3.5 Sinusoidal functions

Given f(\theta) = \sin(2\theta), write the corresponding cosine function.

To convert a sine function to its corresponding cosine function, we use the phase shift relationship, \sin(\theta) = \cos\left(\theta - \dfrac{\pi}{2}\right).

To illustrate this more clearly let's make the substitution x=2 \theta and write the identity as \sin(x)=\cos\left(x-\dfrac{\pi}{2}\right).

Reversing our substitution, we get \sin(2\theta)=\cos\left(2\theta-\dfrac{\pi}{2}\right).

Given f(\theta) = \sin(\theta - \dfrac{\pi}{3}), write the corresponding cosine function.

To convert a sine function to its corresponding cosine function, we use the phase shift relationship, \sin(\theta) = \cos\left(\theta - \dfrac{\pi}{2}\right).

To illustrate this more clearly let's make the substitution x=\theta-\dfrac{\pi}{3} and write the identity as \sin(x)=\cos\left(x-\dfrac{\pi}{2}\right).

Reversing our substitution, we get \sin\left(\theta-\dfrac{\pi}{3}\right)=\cos\left(\theta-\dfrac{\pi}{3}-\dfrac{\pi}{2}\right).

Simplifying the expression gives us:

\sin\left(\theta-\dfrac{\pi}{3}\right) = \cos\left(\theta - \dfrac{5\pi}{6}\right)

Given g(\theta) = \cos\left(3\theta + \dfrac{\pi}{4}\right), write the corresponding sine function.

To convert a sine function to its corresponding cosine function, we use the phase shift relationship, \cos(\theta) = \sin\left(\theta + \dfrac{\pi}{2}\right).

To illustrate this more clearly let's make the substitution x=3\theta + \dfrac{\pi}{4} and write the identity as \cos(x) = \sin\left(x + \dfrac{\pi}{2}\right).

Reversing our substitution, we get \cos\left(3\theta + \dfrac{\pi}{4}\right)=\sin\left(3\theta + \dfrac{\pi}{4} + \dfrac{\pi}{2}\right).

Simplifying the expression gives us:

\cos\left(3\theta + \dfrac{\pi}{4}\right) = \sin\left(3\theta + \dfrac{3\pi}{4}\right)

Given g(\theta) = \cos\left(\theta - \dfrac{\pi}{6}\right), write the corresponding sine function.

To convert a sine function to its corresponding cosine function, we use the phase shift relationship, \cos(\theta) = \sin\left(\theta + \dfrac{\pi}{2}\right).

To illustrate this more clearly let's make the substitution x=\theta - \dfrac{\pi}{6} and write the identity as \cos(x) = \sin\left(x + \dfrac{\pi}{2}\right).

Reversing our substitution, we get \cos\left(\theta - \dfrac{\pi}{6}\right)=\sin\left(\theta - \dfrac{\pi}{6} + \dfrac{\pi}{2}\right).

Simplifying this expression results in:

\cos\left(\theta - \dfrac{\pi}{6}\right) = \sin\left(\theta + \dfrac{\pi}{3}\right)

What do the values of \theta from - 2 \pi \leq \theta \leq 0 represent on f \left( \theta \right) = \sin \theta and the unit circle?

Use the same approach to graphing f \left( \theta \right) = \sin \theta from -2 \pi to 0 from part 1 (b). Recall that negative values of an angle on the coordinate grid represent clockwise rotations around the origin.

The graph of f \left( \theta \right) = \sin \theta between -2 \pi and 0 represents one full clockwise rotation along the unit circle.

Describe the interval on the sine function between 90 \degree and 270 \degree. Explain how this relates to rotations in the unit circle.

90 \degree and 270 \degree are located at \dfrac{\pi}{2} and \dfrac{3 \pi}{2}. Highlight this interval on the graph of f \left(\theta \right) = \sin \theta.

The interval on the unit circle from \dfrac{\pi}{2} to \dfrac{3 \pi}{2} can be highlighted to show the interval that corresponds to the graph of f \left( \theta \right) = \sin \theta.

The graph of sine is decreasing from 90 \degree to 270 \degree while the y-coordinates are also decreasing over the same interval on the unit circle as we rotate counterclockwise.

The graph of sine has a positive interval from 90 \degree to 180 \degree. As the central angle of the unit circle rotates counterclockwise from 90 \degree to 180 \degree, the y-coordinates on the unit circle will start at 1 and decrease to 0. This is why we see a positive interval from 90 \degree to 180 \degree in the graph of the sine function.

The graph of sine has a negative interval from 180 \degree to 270 \degree, and the y-coordinates on the unit circle are also negative over the same interval. As the angle continues to rotate from 180 \degree to 270 \degree, the angle is in the third quadrant, where the y-coordinates are negative, and decrease from 0 to -1.

As we move along the x-axis in the positive direction for the graph of sine, we are rotating counterclockwise along the unit circle. Each value of \theta on the graph corresponds to the angle of \theta on the unit circle.

The amplitude of the sine function is 1. Explain how this relates to the unit circle.

Use the definition of amplitude that it is half the distance between the maximum and minimum values of a periodic function.

Graphically, we see that the sine function reaches a maximum of 1 and a minimum of -1, making its amplitude 1. The amplitude of the sine function is equivalent to the radius of the unit circle, which is 1 unit.

Consider the point \left( \dfrac{\pi}{4}, \dfrac{1}{\sqrt{2}} \right) on the graph. Then, look at the point on the function where \theta is 2 \pi units greater and where \theta is 2 \pi units smaller. What do you notice? Explain why this occurs.

Find the point \left( \dfrac{\pi}{4}, \dfrac{1}{\sqrt{2}} \right) on the graph and \theta \pm 2 \pi.

The points on the function where \theta is 2 \pi units greater or smaller share the same y-value as \dfrac{\pi}{4}.

This makes sense because we know that the sine function is periodic, and its period is 2 \pi, meaning that the graph will continue to repeat the same pattern every multiple of 2 \pi, and if plotted on the unit circle, each of these angles would be coterminal to one another.

State the domain and range.

The graph continues in both the positive and negative x-direction, and reaches a minimum and maximum value for y. Use interval notation to write the domain and range.

Domain: \left( -\infty, \infty \right)

Range: [-3, 1]

Identify the y-intercept.

Just like the parent sine function, there is one y-intercept.

The y-intercept occurs at y=-1.

What is the period?

The period of the function is \pi.

Unlike the parent function of sine, the y-values for the transformed graph begin to repeat after a shorter period.

What is the frequency?

The frequency of the function is the reciprocal of the period.

State the amplitude and midline.

Use the formulas for amplitude and midline to calculate the key features:\text{amplitude}=\dfrac{1}{2} \left( \text{maximum value} - \text{minimum value} \right)

\text{midline: } y=\dfrac{1}{2} \left( \text{maximum value} + \text{minimum value} \right)

If we draw a line at y=-1, we can see that it is exactly halfway between the minimum and maximum points. We can also visually confirm on the graph that the distance between the midline and the maximum and between the midline and the minimum is the amplitude, 2.

Complete the table with values in exact form:

Since the function f \left( \theta \right) = \cos \theta is represented by the x-coordinate on the unit circle, we can use the x-coordinates at each angle on the unit circle to complete the table of values.

Sketch a graph for f \left( \theta \right) = \cos \theta on the domain [-2 \pi, 2 \pi].

Plot the points for [0, 2 \pi] using the table.

Then, we can work backwards along the unit circle to each negative value of \theta which is coterminal to a positive angle between 0 and 2 \pi and match the values of cosine from the unit circle.

For the function from -2\pi to 0, \theta= \dfrac{-\pi}{2} is coterminal to \alpha= \dfrac{3 \pi}{2}, and since \cos \left( \dfrac{3\pi}{2} \right)=0, \cos \left( \dfrac{-\pi}{2} \right) = 0 also. The following coterminal angles give the exact values:

\cos \left( -\pi \right) = \cos \left( \pi \right) = -1

\cos \left( \dfrac{-3 \pi}{2} \right) = \cos \left( \dfrac{\pi}{2} \right) = 0

\cos \left( -2 \pi \right) = \cos \left( 2 \pi \right) = 1

State the sign of \cos \left( \dfrac{-\pi}{12} \right).

Using a revolution of the coordinate plane split into 24 equal parts, we can see that \dfrac{-\pi}{12} is in the fourth quadrant. We can use this to determine the sign of cosine at this angle.

Since the sign of the x-coordinate is positive in the fourth quadrant, \cos \left( \dfrac{- \pi}{12} \right) is positive.

What do the values of \theta from - 2 \pi \leq \theta \leq 0 represent on f \left( \theta \right) = \cos \theta and the unit circle?

Use the same approach to graphing f \left( \theta \right) = \cos \theta from -2 \pi to 0 from part 1 (b). Recall that negative values of an angle on the coordinate grid represent clockwise rotations around the origin.

The graph of f \left( \theta \right) = \cos \theta between -2 \pi and 0 represents one full clockwise rotation along the unit circle.

Describe the interval on the cosine function between 90 \degree and 270 \degree. Explain how this relates to rotations in the unit circle.

90 \degree and 270 \degree are located at \dfrac{\pi}{2} and \dfrac{3 \pi}{2}. Highlight this interval on the graph of f \left(\theta \right) = \cos \theta.

The interval on the unit circle from \dfrac{\pi}{2} to \dfrac{3 \pi}{2} can be highlighted to show the interval that corresponds to the graph of f \left(\theta \right) = \cos \theta.

The graph of cosine is negative from 90 \degree to 270 \degree while the x-coordinates are also negative over the same interval on the unit circle as we rotate counterclockwise.

The graph of cosine has a decreasing interval from 90 \degree to 180 \degree. As the central angle of the unit circle rotates counterclockwise from 90 \degree to 180 \degree, the x-coordinates on the unit circle will start at 0 and decrease to -1. This is why we see a decreasing interval from 90 \degree to 180 \degree in the graph of the cosine function.

The graph of cosine has an increasing interval from 180 \degree to 270 \degree, and the x-coordinates on the unit circle are also increasing over the same interval. As the angle continues to rotate from 180 \degree to 270 \degree, the angle is in the third quadrant, where the x-coordinates increase from -1 to 0.

As we move along the x-axis in the positive direction for the graph of cosine, we are rotating counterclockwise along the unit circle. Each value of \theta on the graph corresponds to the angle of \theta on the unit circle.

The amplitude of the cosine function is 1. Explain how this relates to the unit circle.

Use the definition of amplitude that it is half the distance between the maximum and minimum values of a periodic function.

Graphically, we see that the cosine function reaches a maximum of 1 and a minimum of -1, making its amplitude 1. The amplitude of the cosine function is equivalent to the radius of the unit circle, which is 1 unit.

Consider the point \left( \dfrac{\pi}{3}, \dfrac{1}{2} \right) on the graph. Then, look at the point on the function where \theta is 2 \pi units greater and where \theta is 2 \pi units smaller. What do you notice? Explain why this occurs.

Find the point \left( \dfrac{\pi}{3}, \dfrac{1}{2} \right) on the graph and \theta \pm 2 \pi.

The points on the function where \theta is 2 \pi units greater or smaller share the same y-value as \dfrac{\pi}{3}.

This makes sense because we know that the cosine function is periodic, and its period is 2 \pi, meaning that the graph will continue to repeat the same pattern every multiple of 2 \pi, and if plotted on the unit circle, each of these angles would be coterminal to one another.