Graphs and Inverses of Trigonometric Functions - Pre-Calculus

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Question

Given $F(t)=sin\left(\frac{x}{2}-4\right)$ , what is the period for the function?

Answer

The formula for the period of a sine/cosine function is $\frac{2\pi}{|B|}$ .

With the standard form being:

Since $B=\frac{1}{2}$ , the formula becomes $\frac{2\pi}{\frac{1}{2}}$ .

Simplified, the period is $4\pi$ .

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Question

What could be the function for the following graph?

Answer

What could be the function for the following graph?

Begin by realizing we are dealing with a periodic function, so sine and cosine are your best bet.

Next, note that the range of the function is $-2\leq y \leq 6$ and that the function goes through the point .

From this information, we can find the amplitude:

$A=\frac{6--2}{2}=\frac{8}{2}=4$

So our function must have a out in front.

Also, from the point , we can deduce that the function has a vertical translation of positive two.

The only remaining obstacle, is whether the function is sine or cosine. Recall that sine passes through , while cosine passes through . this means that our function must be a sine function, because in order to be a cosien graph, we would need a horizontal translation as well.

Thus, our answer is:

$y=4\sin(x)+2$

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Question

What is the period of this sine graph?

Answer

The graph has 3 waves between 0 and $2\pi$ , meaning that the length of each of the waves is $2\pi$ divided by 3, or $\frac{2}{3}\pi$ .

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Question

What is the period of this graph?

Answer

One wave of the graph goes exactly from 0 to $2\pi$ before repeating itself. This means that the period is $2\pi$ .

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Question

Please choose the best answer from the following choices.

Find the period of the following function.

Answer

The period is defined as the length of one wave of the function. In this case, one full wave is 180 degrees or $\pi$ radians. You can figure this out without looking at a graph by dividing $2\pi$ with the frequency, which in this case, is 2.

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Question

Please choose the best answer from the following choices.

Find the period of the following function in radians:

Answer

If you look at a graph, you can see that the period (length of one wave) is $2\pi$ . Without the graph, you can divide $2\pi$ with the frequency, which in this case, is 1.

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Question

Write the equation for a cosine graph with a minimum at $\left(\frac{ 3 \pi }{ 8 } , -5\right)$ and a maximum at $\left(\frac{ 11 \pi }{ 8 } , -1 \right)$ .

Answer

The equation for this graph will be in the form $y = A \cos (f (x - h)) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

To write this equation, it is helpful to sketch a graph:

From sketching the maximum and the minimum, we can see that the graph is centered at and has an amplitude of 2.

The distance between the maximum and the minimum is half the wavelength. Here, it is $\frac{ 11 \pi }{ 8 } - \frac{ 3 \pi }{ 8 } = \frac{ 8 \pi }{ 8 } = \pi$ . That means that the full wavelength is $2 \pi$ , so the frequency is 1.

The minimum occurs in the middle of the graph, so to figure out where it starts, subtract $\pi$ from the minimum's x-coordinate:

$\frac{ 3 \pi }{ 8 } - \pi = \frac{ 3 \pi }{8 } - \frac{ 8 \pi }{ 8 } = \frac{ -5 \pi }{8 }$

This graph's equation is

$y = 2 \cos (x + \frac{ 5 \pi }{8 }) -3$ .

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Question

Give the period and frequency for the equation $y = 4 \sin (6x - \pi ) + 2$ .

Answer

Our equation is in the form $y = A \cdot \sin ( fx - h ) + k$

where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

We can look at the equation and see that the frequency, , is .

The period is $\frac{ 2 \pi }{f }$ , so in this case $\frac{ 2 \pi }{6} = \frac{ \pi }{3}$ .

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Question

What is the period of the graph $y = -2 \cos (\frac { \pi }{8 } x + 1 )$ ?

Answer

The equation for this function is in the form $y = A \cos (fx - h) + k$

where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

By looking at the equation, we can see that the frequency, , is $\frac{ \pi }{ 8 }$ .

The period is $\frac{ 2 \pi }{f }$ , so in this case $\frac{ 2 \pi }{ \frac{ \pi }{ 8 }} = 2 \pi \div \frac{ \pi }{ 8 } = \frac{ 2 \pi }{1} \cdot \frac{ 8}{\pi } = 16$ .

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Question

Find angle A of the following triangle:

Answer

We are given the hypotenuse and the side opposite of the angle in question. The trig function that relates these two sides is SIN. Therefore, we can write:

$sin(A) = \frac{12}{22}$

In order to solve for A, we need to take the inverse sin of both sides:

$sin^{-1}(sin(A)) = sin^{-1}(\frac{12}{22})$

which becomes

$A = sin^{-1}\bigg(\frac{12}{22}\bigg) = 33^{\circ}$

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Question

Consider $sin(\theta)=-\frac{1}{2}$ , where theta is valid from $\left [ 0,2\pi\right ]$ . What is a possible value of theta?

Answer

Solve for theta by taking the inverse sine of both sides.

$\theta=sin^-^1\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$

Since this angle is not valid for the given interval of theta, add $2\pi$ radians to this angle to get a valid answer in the interval.

$-\frac{\pi}{6}+2\pi = -\frac{\pi}{6}+\frac{12\pi}{6}= \frac{11\pi}{6}$

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Question

Evaluate:

$cos^-^1\left(-\frac{\sqrt3}{2}\right)+cos^-^1\left(\frac{\sqrt3}{2}\right)$

Answer

First evaluate $cos^-^1\left(-\frac{\sqrt3}{2}\right)$ .

To evaluate inverse cosine, it is necessary to know the domain and range of inverse cosine.

For: $\theta=cos^-^1(x)$

The domain is only valid from .

$\theta$ is only valid from $[0,\pi]$ .

The part is asking for the angle where the x-value of the coordinate is $-\frac{\sqrt3}{2}$ . The only possibility on the unit circle is the second quadrant.

$cos^-^1\left(-\frac{\sqrt3}{2}\right)= \frac{5}{6}\pi$

Next, evaluate $cos^-^1\left(\frac{\sqrt3}{2}\right)$ .

Using the same domain and range restrictions, the only valid angle for the given x-value is in the first quadrant on the unit circle.

$cos^-^1\left(\frac{\sqrt3}{2}\right)= \frac{\pi}{6}$

Therefore:

$cos^-^1\left(-\frac{\sqrt3}{2}\right)+cos^-^1\left(\frac{\sqrt3}{2}\right) = \frac{5}{6}\pi + \frac{\pi}{6}=\frac{6\pi}{6}=\pi$

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Question

Evaluate:

$cos^-^1\left(\frac{1}{2}\right)$

Answer

To find the correct value of $cos^-^1\left(\frac{1}{2}\right)$ , it is necessary to know the domain and range of inverse cosine.

Domain:

Range: $[0,\pi]$

The question is asking for the specific angle when the x-coordinate is half.

The only possibility is located in the first quadrant, and the point of the special angle is $\left(\frac{1}{2},\frac{\sqrt3}{2}\right)$

The special angle for this coordinate is $\frac{\pi}{3}$ .

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Question

Find the value of $\cos^-^1\left(\frac{1}{2}\right)$ .

Answer

In order to determine the value or values of $cos^-^1(\frac{1}{2})$ , it is necessary to know the domain and range of the inverse sine function.

Domain:

Range: $[0,\pi]$

The question is asking for the angle value of theta where the x-value is $\frac{1}{2}$ under the range restriction. Since $x=\frac{1}{2}$ is located in the first and fourth quadrants, the range restriction makes theta only allowable from $[0,\pi]$ . Therefore, the theta value must only be in the first quadrant.

The value of the angle when the x-value is $\frac{1}{2}$ is degrees.

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Question

Find the inverse of the function

.

Make sure the final notation is only in the forms including , , and .

Answer

The easiest way to solve this problem is to simplify the original expression.

$y=csc(x)tan(x)=\frac{1}{sin(x)}\frac{sin(x)}{cos(x)}= \frac{1}{cos(x)}$

$y=\frac{1}{cos(x)}$

To find its inverse, let's exchange and ,

$x=\frac{1}{cos(y)}$

Solving for

$\frac{1}{x}=cos(y)$

$arccos(\frac{1}{x})=y$

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Question

Which of the given functions has the greatest amplitude?

Answer

The amplitude of a function is the amount by which the graph of the function travels above and below its midline. When graphing a sine function, the value of the amplitude is equivalent to the value of the coefficient of the sine. Similarly, the coefficient associated with the x-value is related to the function's period. The largest coefficient associated with the sine in the provided functions is 2; therefore the correct answer is $2\sin(x)$ .

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

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Question

What is the amplitude of $y=3\sin\left(\frac{x}{5}\right)}$ ?

Answer

For any equation in the form $y=a\sin(bx)$ , the amplitude of the function is equal to $\left | a\right |$ .

In this case, and $b=\frac{1}{5}$ , so our amplitude is .

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Question

What is the amplitude of ?

Answer

The formula for the amplitude of a sine function is from the form:

.

In our function, .

Therefore, the amplitude for this function is .

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Question

Find the amplitude of the following trig function: $y=-2-3sin(2\theta-5)$

Answer

Rewrite $y=-2-3sin(2\theta-5)$ so that it is in the form of:

$y=-3sin(2\theta-5)-2$

The absolute value of is the value of the amplitude.

$\textup{Amplitude= }\left | A\right |= \left | -3\right |=3$

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Question

Find the amplitude of the function.

Answer

For the sine function

where $A\epsilon , \mathbb{R}$

the amplitude is given as $\left | A\right |$ .

As such the amplitude for the given function

is

$\left | 5\right |=5$ .

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