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# Definition of the Inverse of Trigonometric Ratios

## Solving for an angle given two sides of a right triangle.

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Definition of the Inverse of Trigonometric Ratios

You are in band practice one day when something catches your eye. One of your fellow students has a musical instrument that you haven't seen played before - a triangle. This is an instrument that is a metal triangle that the musician plays by striking it. Walking over to your classmate, you ask him about the instrument. During a conversation about the triangle, you wonder if it would be possible to make different types of triangle instruments to make different sounds.

You ask your science teacher about this, and she is excited that you have taken an interest in the topic. Together you decide to devise some new instruments based on different length sides of triangles, and then try them out to see how they sound. To begin, your teacher asks you to create a list of a few different sides of triangle lengths, and then list the interior angles of the triangles you come up with.

You immediately get to work and generate a list. The first triangle has sides of 12 cm, 35 cm, and 37 cm, and is a right triangle. However, you realize that you aren't sure of the interior angles of the triangles. You are about to get out a paper and pencil to start plotting the triangles when you start to wonder if there might be an easier way to use math in finding the measurement of the angles instead of plotting them and measuring by hand.

### Inverse of Trigonometric Ratios

In this section, we'll discuss and apply examples of the inverse functions for trig ratios.

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

sinθ=yrsin1yr=θcosθ=xrcos1xr=θtanθ=yxtan1yx=θcotθ=xycot1xy=θcscθ=rycsc1ry=θsecθ=rxsec1rx=θ\begin{align*}\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\end{align*}

These ratios can be used to find any θ\begin{align*}\theta\end{align*} in standard position or in a triangle.

In a sense, this is a way of ‘‘undoing’’ a trig function. Before, to find a trig function, you would use the ratio of two sides. Now, by using the inverse trig ratio, you can find angles when you need them.

Let's look at some problems involving inverse trigonometric ratios.

1. Find the measure of the angles below.

You need to use the sine function, and then find the inverse of sine.

sinx=725sin1725=xx=16.26\begin{align*}\sin x=\frac{7}{25} \rightarrow \sin^{-1} \frac{7}{25} = x \rightarrow x = 16.26^\circ\end{align*}

You need to use the tangent function, and then find the inverse of tangent.

tanx=409tan1409=xx=77.32\begin{align*}\tan x=\frac{40}{9} \rightarrow \tan^{-1} \frac{40}{9} = x \rightarrow x = 77.32^\circ\end{align*}

The trigonometric value tanθ=409\begin{align*}\tan \theta = \frac{40}{9}\end{align*} 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. 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.

2. Find the angle, θ\begin{align*}\theta\end{align*}, in standard position.

The tanθ=yx\begin{align*}\tan \theta = \frac{y}{x}\end{align*} or, in this case, tanθ=811\begin{align*}\tan \theta = \frac{8}{-11}\end{align*}. Using the inverse tangent, you get tan1811=36.03\begin{align*}\tan^{-1} -\frac{8}{11} = -36.03^\circ\end{align*}. This means that the reference angle is 36.03\begin{align*}36.03^\circ\end{align*}. This value of 36.03\begin{align*}36.03^\circ\end{align*} 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\begin{align*}36.03^\circ\end{align*} from 180\begin{align*}180^\circ\end{align*}, yielding 143.97\begin{align*}143.97^\circ\end{align*}.

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

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?

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.

tanθ=2560tanθ=0.4166tan1(tanθ)=tan1(0.4166)θ=22.6\begin{align*}\tan \theta = \frac{25}{60}\\ \tan \theta = 0.4166\\ \tan^{-1}(\tan \theta) = \tan^{-1}(0.4166)\\ \theta = 22.6^\circ\end{align*}

The angle of depression at which the light must be placed to light the rink is 22.6\begin{align*}22.6^\circ\end{align*}

### Examples

#### Example 1

Earlier, you were given a problem on inverse trig functions.

Since you now know about inverse trigonometric ratios, you know that you can apply the inverse of a trig function to help solve this problem. For example, it is straightforward to apply the tangent function:

tanθ=3512θ=tan13512θ=71.08\begin{align*} \tan \theta = \frac{35}{12}\\ \theta = \tan^{-1} \frac{35}{12}\\ \theta = 71.08^\circ\\ \end{align*}

You can find the other angle the in a similar manner, this time using the sine function:

sinθ=1237θ=sin11237θ=18.92\begin{align*} \sin \theta = \frac{12}{37}\\ \theta = \sin^{-1} \frac{12}{37}\\ \theta = 18.92^\circ\\ \end{align*}

#### Example 2

Find the value of the missing angle.

cosθ=1217cos11217=45.1\begin{align*}\cos \theta = \frac{12}{17} \rightarrow \cos^{-1} \frac{12}{17} = 45.1^\circ\end{align*}

#### Example 3

Find the value of the missing angle.

sinθ=3136sin13136=59.44\begin{align*}\sin \theta = \frac{31}{36} \rightarrow \sin^{-1} \frac{31}{36} = 59.44^\circ\end{align*}

#### Example 4

What is the value of the angle with its terminal side passing through (-14, -23)?

This problem uses tangent inverse. tanx=2314x=tan12314=58.67\begin{align*}\tan x = \frac{-23}{-14} \rightarrow x = \tan^{-1} \frac{23}{14} = 58.67^\circ\end{align*} (value graphing calculator will produce). However, this is the reference angle. Our angle is in the third quadrant because both the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values are negative. The angle is 180+58.67=238.67\begin{align*}180^\circ + 58.67^\circ = 238.67^\circ\end{align*}.

### Review

Find the measure of angle A in each triangle below.

Use inverse tangent to find the value of the angle with its terminal side passing through each of the given points.

1. (-3,4)
2. (12,13)
3. (-4, -7)
4. (5, -4)
5. (-6, 10)
6. (-3, -10)
7. (6, 8)

Use inverse trigonometry to solve each problem.

1. A 30 foot building casts a 60 foot shadow. What is the angle that the sun hits the building?
2. Over 3 miles (horizontal), a road rises 100 feet (vertical). What is the angle of elevation?
3. An 80 foot building casts a 20 foot shadow. What is the angle that the sun hits the building?

To see the Review answers, open this PDF file and look for section 4.1.

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