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# 4.1: Basic Inverse Trigonometric Functions

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## Introduction

Recall that an inverse function is a reflection of the function over the line $y = x$. In order to find the inverse of a function, you must switch the $x$ and $y$ values and then solve for $y$. A function has an inverse if and only if it has exactly one output for every input and exactly one input for every output. All of the trig functions fit these criteria over a specific range. In this chapter, we will explore inverse trig functions and equations.

## Learning Objectives

• Understand and evaluate inverse trigonometric functions.
• Extend the inverse trigonometric functions to include the $\csc^{-1}, \sec^{-1}$ and $\cot^{-1}$ functions.
• Apply inverse trigonometric functions to the critical values on the unit circle.

## Defining the Inverse of the Trigonometric Ratios

Recall from Chapter 1, the ratios of the six trig functions and their inverses, with regard to the unit circle.

$\sin \theta &= \frac{y}{r} \rightarrow \sin^{-1} \frac{y}{r} = \theta && \cos \theta = \frac{x}{r} \rightarrow \cos^{-1} \frac{x}{r} = \theta\\\tan \theta &= \frac{y}{x} \rightarrow \tan^{-1} \frac{y}{x} = \theta && \cot \theta = \frac{x}{y} \rightarrow \cot^{-1} \frac{x}{y} = \theta\\\csc \theta &= \frac{r}{y} \rightarrow \csc^{-1} \frac{r}{y} = \theta && \sec \theta = \frac{r}{x} \rightarrow \sec^{-1} \frac{r}{x} = \theta$

These ratios can be used to find any $\theta$ in standard position or in a triangle.

Example 1: Find the measure of the angles below.

a.

b.

Solution: For part a, you need to use the sine function and part b utilizes the tangent function. Because both problems require you to solve for an angle, the inverse of each must be used.

a. $\sin x=\frac{7}{25} \rightarrow \sin^{-1} \frac{7}{25} = x \rightarrow x = 16.26^\circ$

b. $\tan x=\frac{40}{9} \rightarrow \tan^{-1} \frac{40}{9} = x \rightarrow x = 77.32^\circ$

The trigonometric value $\tan \theta = \frac{40}{9}$ of the angle is known, but not the angle. In this case the inverse of the trigonometric function must be used to determine the measure of the angle. (Directions for how to find inverse function values in the graphing calculator are in Chapter 1). The inverse of the tangent function is read “tangent inverse” and is also called the arctangent relation. The inverse of the cosine function is read “cosine inverse” and is also called the arccosine relation. The inverse of the sine function is read “sine inverse” and is also called the arcsine relation.

Example 2: Find the angle, $\theta$, in standard position.

Solution: The $\tan \theta = \frac{y}{x}$ or, in this case, $\tan \theta = \frac{8}{-11}$. Using the inverse tangent, you get $\tan^{-1} -\frac{8}{11} = -36.03^\circ$. This means that the reference angle is $36.03^\circ$. This value of $36.03^\circ$ is the angle you also see if you move counterclockwise from the -x axis. To find the corresponding angle in the second quadrant (which is the same as though you started at the +x axis and moved counterclockwise), subtract $36.03^\circ$ from $180^\circ$, yielding $143.97^\circ$.

Recall that inverse trigonometric functions are also used to find the angle of depression or elevation.

Example 3: A new outdoor skating rink has just been installed outside a local community center. A light is mounted on a pole 25 feet above the ground. The light must be placed at an angle so that it will illuminate the end of the skating rink. If the end of the rink is 60 feet from the pole, at what angle of depression should the light be installed?

Solution: In this diagram, the angle of depression, which is located outside of the triangle, is not known. Recall, the angle of depression equals the angle of elevation. For the angle of elevation, the pole where the light is located is the opposite and is 25 feet high. The length of the rink is the adjacent side and is 60 feet in length. To calculate the measure of the angle of elevation the trigonometric ratio for tangent can be applied.

$\tan \theta & = \frac{25}{60}\\\tan \theta & = 0.4166\\\tan^{-1}(\tan \theta) & = \tan^{-1}(0.4166)\\\theta & = 22.6^\circ$

The angle of depression at which the light must be placed to light the rink is $22.6^\circ$

## Exact Values for Inverse Sine, Cosine, and Tangent

Recall the unit circle and the critical values. With the inverse trigonometric functions, you can find the angle value (in either radians or degrees) when given the ratio and function. Make sure that you find all solutions within the given interval.

Example 4: Find the exact value of each expression without a calculator, in $[0, 2\pi)$.

a. $\sin^{-1} \left ( -\frac{\sqrt{3}}{2} \right )$

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

c. $\tan^{-1} \sqrt{3}$

Solution: These are all values from the special right triangles and the unit circle.

a. Recall that $-\frac{\sqrt{3}}{2}$ is from the $30-60-90$ triangle. The reference angle for $\sin$ and $\frac{\sqrt{3}}{2}$ would be $60^\circ$. Because this is sine and it is negative, it must be in the third or fourth quadrant. The answer is either $\frac{4\pi}{3}$ or $\frac{5\pi}{3}$.

b. $-\frac{\sqrt{2}}{2}$ is from an isosceles right triangle. The reference angle is then $45^\circ$. Because this is cosine and negative, the angle must be in either the second or third quadrant. The answer is either $\frac{3\pi}{4}$ or $\frac{5\pi}{4}$.

c. $\sqrt{3}$ is also from a $30-60-90$ triangle. Tangent is $\sqrt{3}$ for the reference angle $60^\circ$. Tangent is positive in the first and third quadrants, so the answer would be $\frac{\pi}{3}$ or $\frac{4\pi}{3}$.

Notice how each one of these examples yield two answers. This poses a problem when finding a singular inverse for each of the trig functions. Therefore, we need to restrict the domain in which the inverses can be found, which will be addressed in the next section. Unless otherwise stated, all angles are in radians.

## Finding Inverses Algebraically

In the Prerequisite Chapter, you learned that each function has an inverse relation and that this inverse relation is a function only if the original function is one-to-one. A function is one-to-one when its graph passes both the vertical and the horizontal line test. This means that every vertical and horizontal line will intersect the graph in exactly one place.

This is the graph of $f(x) = \frac{x}{x+1}$. The graph suggests that $f$ is one-to-one because it passes both the vertical and the horizontal line tests. To find the inverse of $f$, switch the $x$ and $y$ and solve for $y$.

First, switch $x$ and $y$.

$x = \frac{y}{y+1}$

Next, multiply both sides by $(y + 1)$.

$(y+1)x & = \frac{y}{y+1} (y+1)\\x(y+1) & = y$

Then, apply the distributive property and put all the $y$ terms on one side so you can pull out the $y$.

$xy+x & = y\\xy-y & = -x\\y(x-1) & = -x$

Divide by $(x - 1)$ to get $y$ by itself.

$y = \frac{-x}{x-1}$

Finally, multiply the right side by $\frac{-1}{-1}$.

$y = \frac{x}{1-x}$

Therefore the inverse of $f$ is $f^{-1}(x)=\frac{x}{1-x}$.

The symbol $f^{-1}$ is read “$f$ inverse” and is not the reciprocal of $f$.

Example 5: Find the inverse of $f(x) = \frac{1}{x-5}$ algebraically.

Solution: To find the inverse algebraically, switch $f(x)$ to $y$ and then switch $x$ and $y$.

$y & =\frac{1}{x-5}\\x & = \frac{1}{y-5}\\x(y-5) & = 1\\xy-5x & = 1\\xy & = 5x+1\\y & = \frac{5x+1}{x}$

Example 6: Find the inverse of $f(x) = 5 \sin^{-1} \left ( \frac{2}{x-3} \right )$

Solution:

a. $f(x) & = 5 \sin^{-1} \left ( \frac{2}{x-3} \right )\\x & = 5 \sin^{-1} \left ( \frac{2}{y-3} \right )\\\frac{x}{5} & = \sin^{-1} \left ( \frac{2}{y-3} \right )\\\sin \frac{x}{5} & = \left ( \frac{2}{y-3} \right )\\(y-3)\sin \frac{x}{5} & = 2\\(y-3) & = \frac{2}{\sin \frac{x}{5} }\\y & = \frac{2}{\sin \frac{x}{5} } + 3$

Example 7: Find the inverse of the trigonometric function $f(x) = 4 \tan^{-1}(3x + 4)$

Solution:

$x & = 4\tan^{-1}(3y+4)\\\frac{x}{4} & = \tan^{-1}(3y+4)\\\tan \frac{x}{4} & = 3y+4\\\tan \frac{x}{4} - 4 & = 3y\\\frac{\tan \frac{x}{4} -4}{3} & = y\\f^{-1} (x) & = \frac{\tan \frac{x}{4} -4}{3}$

## Points to Consider

• What is the difference between an inverse and a reciprocal?
• Considering that most graphing calculators do not have $\csc, \sec$ or $\cot$ buttons, how would you find the inverse of each of these?
• Besides algebraically, is there another way to find the inverse?

## Review Questions

1. Use the special triangles or the unit circle to evaluate each of the following:
1. $\cos 120^\circ$
2. $\csc \frac{3\pi}{4}$
3. $\tan \frac{5\pi}{3}$
2. Find the exact value of each inverse function, without a calculator in $[0, 2\pi)$:
1. $\cos^{-1}(0)$
2. $\tan^{-1} \left ( -\sqrt{3} \right )$
3. $\sin^{-1}\left ( -\frac{1}{2} \right )$

Find the value of the missing angle.

1. What is the value of the angle with its terminal side passing through (-14, -23)?
2. A 9-foot ladder is leaning against a wall. If the foot of the ladder is 4 feet from the base of the wall, what angle does the ladder make with the floor?

Find the inverse of the following functions.

1. $f(x) = 2x^3-5$
2. $y = \frac{1}{3}\tan^{-1} \left ( \frac{3}{4}x-5 \right )$
3. $g(x)=2\sin (x-1)+4$
4. $h(x) = 5-\cos^{-1}(2x+3)$

## Review Answers

1. $-\frac{1}{2}$
2. $\sqrt{2}$
3. $-\sqrt{3}$
1. $\frac{\pi}{2}, \frac{3\pi}{2}$
2. $\frac{2\pi}{3}, \frac{5\pi}{3}$
3. $\frac{11\pi}{6}, \frac{7\pi}{6}$
1. $\cos \theta = \frac{12}{17} \rightarrow \cos^{-1} \frac{12}{17} = 45.1^\circ$
2. $\sin \theta = \frac{25}{36} \rightarrow \sin^{-1} \frac{31}{36} = 59.44^\circ$
3. This problem uses tangent inverse. $\tan x = \frac{-23}{-14} \rightarrow x = \tan^{-1} \frac{23}{14} = 58.67^\circ$ (value graphing calculator will produce). However, this is the reference angle. Our angle is in the third quadrant because both the $x$ and $y$ values are negative. The angle is $180^\circ + 58.67^\circ = 238.67^\circ$.
4. $\cos A & = \frac{4}{9}\\\cos^{-1} \frac{4}{9} & = A\\\angle{A} & = 63.6^\circ$
5. $f(x) & = 2x^3-5\\y & = 2x^3-5\\x & = 2y^3-5\\x + 5 & = 2y^3\\\frac{x+5}{2} & = y^3\\\sqrt[3]{\frac{x+5}{2}} & = y$
6. $y & = \frac{1}{3} \tan^{-1}\left ( \frac{3}{4}x-5 \right )\\x & = \frac{1}{3}\tan^{-1}\left ( \frac{3}{4}y-5 \right )\\3x & = \tan^{-1}\left ( \frac{3}{4}y-5 \right )\\\tan (3x) & = \frac{3}{4}y-5\\\tan (3x) + 5 & = \frac{3}{4}y\\\frac{4(\tan (3x)+5)}{3} & = y$
7. $g(x) & = 2 \sin (x-1)+4\\y & = 2 \sin (x-1)+4\\x & = 2 \sin (y-1)+4\\x-4 & = 2 \sin (y-1)\\\frac{x-4}{2} & = \sin (y-1)\\\sin^{-1} \left ( \frac{x-4}{2} \right ) & = y-1\\1 + \sin^{-1} \left ( \frac{x-4}{2} \right ) & = y$
8. $h(x) & = 5-\cos^{-1} (2x+3)\\y & = 5-\cos^{-1} (2x+3)\\x & = 5-\cos^{-1} (2y+3)\\x-5 & = -\cos^{-1} (2y+3)\\5-x & = \cos^{-1} (2y+3)\\\cos (5-x) & = 2y+3\\\cos (5-x)-3 & = 2y\\\frac{\cos (5-x)-3}{2} & = y$

Feb 23, 2012

## Last Modified:

Aug 21, 2014
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