# Intro to inverse trig functions

CCSS Math: HSG.SRT.C.8

Learn about arcsine, arccosine, and arctangent, and how they can be used to solve for a missing angle in right triangles.

Let's take a look at a new type of trigonometry problem. Interestingly, these problems can't be solved with sine, cosine, or tangent.

**A problem:**In the triangle below, what is the measure of angle $L$?

**What we know:**Relative to $\angle L$, we know the lengths of the opposite and adjacent sides, so we can write:

But this doesn't help us find the measure of $\angle L$. We're stuck!

**What we need:**We need new mathematical tools to solve problems like these. Our old friends sine, cosine, and tangent aren’t up to the task. They take angles and give side ratios, but we need functions that take side ratios and give angles. We need

**inverse trig functions**!

# The inverse trigonometric functions

We already know about

*inverse operations*. For example, addition and subtraction are inverse operations, and multiplication and division are inverse operations. Each operation does the*opposite*of its inverse.The idea is the same in trigonometry.

*Inverse trig functions*do the opposite of the “regular” trig functions. For example:- Inverse sine $(\sin^{-1})$ does the opposite of the sine.
- Inverse cosine $(\cos^{-1})$ does the opposite of the cosine.
- Inverse tangent $(\tan^{-1})$ does the opposite of the tangent.

In general, if you know the trig ratio but not the angle, you can use the corresponding inverse trig function to find the angle. This is expressed mathematically in the statements below.

Trigonometric functions input angles and output side ratios | Inverse trigonometric functions input side ratios and output angles | |
---|---|---|

$\sin (\theta)=\dfrac {\text{opposite}}{\text{hypotenuse}}$ | $\rightarrow$ | $\sin^{-1}\left(\dfrac {\text{opposite}}{\text{hypotenuse}}\right)=\theta$ |

$\cos (\theta)=\dfrac {\text{adjacent}}{\text{hypotenuse}}$ | $\rightarrow$ | $\cos^{-1}\left(\dfrac {\text{adjacent}}{\text{hypotenuse}}\right)=\theta$ |

$\tan (\theta)=\dfrac {\text{opposite}}{\text{adjacent}}$ | $\rightarrow$ | $\tan^{-1}\left(\dfrac {\text{opposite}}{\text{adjacent}}\right)=\theta$ |

## Misconception alert!

The expression $\sin^{-1}(x)$ is not the same as $\dfrac{1}{\sin(x)}$. In other words, the $\small{-1}$ is not an exponent. Instead, it simply means inverse function.

However, there is an alternate notation that avoids this pitfall! The inverse sine can also be expressed as $\arcsin$, the inverse cosine as $\arccos$, and the inverse tangent as $\arctan$. This notation is common in computer programming languages, but not in mathematics.

# Solving the introductory problem

In the introductory problem, we were given the opposite and adjacent side lengths, so we can use inverse tangent to find the angle.