Trigonometry - Graphs of Inverse Trigonometric Functions

True or False: The inverse of the function $\sin (x)$ is also a function.

True

False

Explanation

Consider the graph of the function . It passes the vertical line test, that is if a vertical line is drawn anywhere on the graph it only passes through a single point of the function. This means that is a function.

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Now, for its inverse to also be a function it must pass the horizontal line test. This means that if a horizontal line is drawn anywhere on the graph it will only pass through one point.

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This is not true, and we can also see that if we graph the inverse of ( $sin^{-1}(x)$ ) that this does not pass the vertical line test and therefore is not a function. If you wish to graph the inverse of , then you must restrict the domain so that your graph will pass the vertical line test.

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Which of the following is the graph of the inverse of with $D: \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ ?

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Explanation

Note that the inverse of is not $\frac{1}{sin(x)}$ , that is the reciprocal. The inverse of is $sin^{-1}(x)$ also written as . The graph of with $D: \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ is as follows.

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And so the inverse of this graph must be the following with $D: -1 \leq x \leq 1$ and $R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

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Which best describes the easiest method to graph an inverse trigonometric function (or any function) based on the parent function?

Let $x = \frac{1}{x}$ so where for the parent function, $y = cos(\frac{1}{x})$ for the inverse function.

Let $y = \frac{1}{y}$ so where for the parent function $y = \frac{1}{cos(x)}$ for the inverse function.

The and values are switched so where for the parent function, for the inverse function.

The values are swapped with $\frac{1}{y}$ so where for the parent function, $x = cos(\frac{1}{y})$ for the inverse function.

Explanation

To find an inverse function you swap the and values. Take for example, to find the inverse we use the following method.

(swap the and values)

$cos^{-1}(x) = y$ (solving for )

$y = cos^{-1}(x)$

Which of the following represents the graph of $tan^{-1}(x)$ with $R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$ ?

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Explanation

If we are looking for the graph of $tan^{-1}(x)$ with $R: \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ , that means this is the inverse of with $D:\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ . The graph of with $D:\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ is

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Switching the and values to graph the inverse we get the graph

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Which of the following is the graph of $cos^{-1}(x) + 2$ with $D: -1 \leq x \leq 1$ ?

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Explanation

We first need to think about the graph of the function $cos^{-1}(x)$ .

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Using the formula where is the vertical shift, we have to perform a transformation of moving the function $cos^{-1}(x)$ up two units on the graph.

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Which of the following is the correct graph and range of the inverse function of with $D:\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ ?

$R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

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$R:-\pi \leq y \leq \pi$

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$R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

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$R:-\pi \leq y \leq \pi$

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Explanation

First, we must solve for the inverse of

$\frac{x}{2} = sin(y)$

$sin^-1(\frac{x}{2}) = y$

So now we are trying to find the range of and plot the function $y = sin^{-1}(\frac{x}{2})$ . Let’s start with the graph of $sin^{-1}(x)$ . We know the domain is .

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Now using the formula $y = D + Acos^{-1}(B(x-C))$ where $\frac{2\pi}{B}$ = Period, the period of $y = sin^{-1}(\frac{x}{2})$ is $4\pi$ . And so we perform a transformation to the graph of $sin^{-1}(x)$ to change the period from $2\pi$ to $4\pi$ .

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We can see that the graph has a range of $[\frac{-\pi}{2},\frac{\pi}{2}]$

True or False: The domain for will always be all real numbers no matter the value of or any transformations applied to the tangent function.

True

False

Explanation

This is true because just as the range of is all real numbers due to the vertical asymptotes of the function, the function extends to all values of but is limited in its values of . No matter the transformations applied, all values of will still be reached.

Which of the following is the graph of $\frac{1}{2}cos^{-1}(x) + 1$ ?

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Explanation

First, we must consider the graph of $cos^{-1}(x)$ .

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Using the formula $D + Acos^{-1}(B(x-C))$ we can apply the transformations step-by-step. First we will transform the amplitude, so $A = \frac{1}{2}$ so we must shorten the amplitude to $\frac{1}{2}$ .