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### Course: High school geometry>Unit 5

Lesson 7: Solving for an angle in a right triangle using the trigonometric ratios

# Intro to inverse trig functions

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 $\mathrm{\angle }L$, we know the lengths of the opposite and adjacent sides, so we can write:
$\mathrm{tan}\left(L\right)=\frac{\text{opposite}}{\text{adjacent}}=\frac{35}{65}$
But this doesn't help us find the measure of $\mathrm{\angle }L$. 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 $\left({\mathrm{sin}}^{-1}\right)$ does the opposite of the sine.
• Inverse cosine $\left({\mathrm{cos}}^{-1}\right)$ does the opposite of the cosine.
• Inverse tangent $\left({\mathrm{tan}}^{-1}\right)$ 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 ratiosInverse trigonometric functions input side ratios and output angles
$\mathrm{sin}\left(\theta \right)=\frac{\text{opposite}}{\text{hypotenuse}}$$\to$${\mathrm{sin}}^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)=\theta$
$\mathrm{cos}\left(\theta \right)=\frac{\text{adjacent}}{\text{hypotenuse}}$$\to$${\mathrm{cos}}^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)=\theta$
$\mathrm{tan}\left(\theta \right)=\frac{\text{opposite}}{\text{adjacent}}$$\to$${\mathrm{tan}}^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)=\theta$

The expression ${\mathrm{sin}}^{-1}\left(x\right)$ is not the same as $\frac{1}{\mathrm{sin}\left(x\right)}$. In other words, the $-1$ is not an exponent. Instead, it simply means inverse function.
FunctionGraph
$\mathrm{sin}\left(x\right)$
${\mathrm{sin}}^{-1}\left(x\right)$ (also called $\mathrm{arcsin}\left(x\right)$) |
$\frac{1}{\mathrm{sin}x}$ (also called $\mathrm{csc}\left(x\right)$) |
However, there is an alternate notation that avoids this pitfall! We can also express the inverse sine as $\mathrm{arcsin}$, the inverse cosine as $\mathrm{arccos}$, and the inverse tangent as $\mathrm{arctan}$. This notation is common in computer programming languages, and less common 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.

## Now let's try some practice problems.

Problem 1
Given $\mathrm{△}KIP$, find $m\mathrm{\angle }I$.
${}^{\circ }$

Problem 2
Given $\mathrm{△}DEF$, find $m\mathrm{\angle }E$.
${}^{\circ }$

Problem 3
Given $\mathrm{△}LYN$, find $m\mathrm{\angle }Y$.
${}^{\circ }$

Challenge problem
Solve the triangle completely. That is, find all unknown sides and unknown angles.
$OE=$
$m\mathrm{\angle }O=$
${}^{\circ }$
$m\mathrm{\angle }Z=$
${}^{\circ }$

## Want to join the conversation?

• this might sound like a silly question, but i was hoping that sin(90) = 2 sin(45).
Why doesn't that work? Trig functions are all about ratios and relations, the least i could expect was to find a relation like that...
• this might have been possible if sin was a linear function which its not....
• Love the site, but slightly thrown having to switch from using DEG mode to RAD mode to get correct answer on inverse trig questions. Would be good to be given a heads-up that this was necessary. And why it was necessary. Which I. Still haven't really figured out!
• DEG mode stands for "degree". This means that your calculator interprets and outputs angles in the unit of degrees. RAD mode stands for "radian". This means that your calculator interprets and outputs angles in the unit of radians. If you are not sure what radians are, I suggest you watch the KA videos on them. Switching between DEG mode and RAD mode on a calculator is similar to switching between "miles per hour" and "kilometers per hour" on a speedometer. You still get the same speeds, but in different units.
Comment if you have questions!
• how to turn calculator on
• If it is not turning on then you need to replace the batteries. Hope this helps.
(:
• How to calculate the inverse function in a calculator?
• Many calculators (TI and others) have the inverse trig funcdtions (sin-1, cos-1, tan-1) on the same button, but using the 2nd sin function. Do not know which particular calculator you are talking about.
• Every time I use one of the formulas, I get a much smaller answer than I’m supposed to. Even in the introductory problem, if I put tan^-1(35/65) into my calculator, I get the answer of ~0.49 when its supposed to be ~28.30. I can’t find a way to get the correct answers, am I missing a step?
• This is common mistake, on your calculator, make sure you are in degree mode, you are currently in radian mode.
• What if we do not want to use a calculator and do it manually?
• Then you will need access to trigonometric tables that you can read in reverse. This is how I used to estimate the inverse trigonometric functions when I was in high school. I still have a book of tables to trig functions, logarithms, and z-scores (among other useful relationships) to which I refer when solving some problems, but the modern method of using a calculator or computer to access this information is usually more efficient and precise.
• could some one explain what ' round your answer to the nearest hundredth degree' means. its mentioned in the second practice question.
• "To the nearest hundredth of a degree" means to solve it, and then round it to 2 decimal places. The first place is tenths, and the second place is hundredths.
Example: Problem 3.
We're trying to find angle Y. We have the adjacent side length and the hypotenuse length. With the sides adjacent and hypotenuse, we can use the Cosine function to determine angle Y.
CosY = 3/10
CosY = 0.30
This is where the Inverse Functions come in. If we know that CosY = 0.30, we're trying to find the angle Y that has a Cosine 0.30. To do so: