## Introduction to Inverse Trig Functions

We studied **Inverses of Functions** here; we remember that getting the inverse of a function is basically switching the $ x$- and $ y$-values and solving for the other variable. The inverse of a function is symmetrical (a mirror image) around the line $ y=x$. Here’s an example of how we’d find an inverse algebraically with a trig function:

(Note that if $ {{\sin }^{-1}}\left( x \right)=y$, then $ \sin \left( y \right)=x$. When we take the inverse of a trig function, what’s in parentheses (the $ x$ here), is **not an angle**, but the actual **sin** (trig) value. The trig inverse (the $ y$) is the angle (usually in radians). Also note that the **–1** is **not an exponent**, so we are not putting anything in a denominator. We can also write trig functions with “arcsin” instead of $ {{\sin }^{-1}}$: if $ \arcsin \left( x \right)=y$, then $ \sin \left( y \right)=x$.)

The same principles apply for the **inverses of six trigonometric functions**, but since the trig functions are periodic (repeating), these functions don’t have inverses, unless we **restrict the domain**. As shown below, we restrict the domains to certain **quadrants** so the original function passes the **horizontal line test** and thus the inverse function passes the **vertical line test**. Thus, the inverse trig functions are **one-to-one** **functions**, meaning every element of the range of the function corresponds to exactly one element of the domain.

As an example, in order to make an inverse trig function an actual function, only take the values between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$, so the **sin** function passes the horizontal line test (meaning its inverse is a function):

To help remember which quadrants the inverse trig functions **on the Unit Circle** will come from, I use these “sun” diagrams:

The inverse **cos**, **sec**, and **cot** functions return values in the **I and II Quadrants** (between **0** and $ 2\pi $), and the inverse **sin**,** csc**,** and tan** functions return values in the **I and IV Quadrants **(between $ -\frac{\pi }{2}$ and $ \frac{\pi }{2}$), with **negative values** in** Quadrant IV**). (I would just memorize these rules, since it’s simple to do so). Again, these are called **domain restrictions** for the inverse trig functions.

**Important Note: **There is a subtle distinction between **finding inverse trig functions** and **solving for trig functions**. If we want $ \displaystyle {{\sin }^{{-1}}}\left( {\frac{{\sqrt{2}}}{2}} \right)$ for example, we only pick the answers from **Quadrants** **I** and **IV**, so we get $ \displaystyle \frac{\pi }{4}$ only. But if we are solving $ \displaystyle \sin \left( x \right)=\frac{{\sqrt{2}}}{2}$ like in the **Solving Trigonometric Functions** section, there are no domain restrictions, so we get $ \displaystyle \frac{\pi }{4}$ and $ \displaystyle \frac{{3\pi }}{4}$ in the interval $ \left( {0,2\pi } \right)$, and $ \displaystyle \frac{\pi }{4}+2\pi k$ and $ \displaystyle \frac{3\pi }{4}+2\pi k$ over all reals.

## Graphs of Inverse Trig Functions

Here are tables of the inverse trig functions and their **t-charts**, **graphs**, **domain**,** range **(also called the** principal interval**), and any **asymptotes**.

## Evaluating Inverse Trig Functions – Special Angles

When you are asked to evaluate inverse functions, you may see the notation $ {{\sin }^{-1}}$ or **arcsin**; they mean the same thing. The following examples use angles that are **special values or special angles**: angles that have trig values that we can compute exactly, since they come right off the **Unit Circle:**

To do these problems, remember again the “sun” diagrams to make sure you’re **getting the angle back from the correct quadrant**: .

When using the **Unit Circle**, when the answer is in **Quadrant IV**, it must be negative (go backwards from the $ (1, 0)$ point). For example, for the $ \displaystyle {{\sin }^{-1}}\left( -\frac{1}{2} \right)$ or $ \displaystyle \arcsin \left( -\frac{1}{2} \right)$, the angle is **330°**, or $ \displaystyle \frac{11\pi }{6}$. But since our answer has to be between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$, we need to change this to the co-terminal angle $ -30{}^\circ $, or $ \displaystyle -\frac{\pi }{6}$.

To get the inverses for the **reciprocal functions**, take the reciprocal of what’s in the parentheses and then use the “normal” trig functions. For example, to get $ {{\sec }^{-1}}\left( -\sqrt{2} \right)$, look for $ \displaystyle {{\cos }^{-1}}\left( -\frac{1}{\sqrt{2}} \right)$, which is $ \displaystyle {{\cos }^{-1}}\left( -\frac{\sqrt{2}}{2} \right)$, which is $ \displaystyle \frac{3\pi }{4}$, or **135°**.

**Check your work**: For all inverse trig functions of a positive argument (given the **correct domain**), we should get an angle in **Quadrant I** ($ \displaystyle 0\le \theta \le \frac{\pi }{2}$). For the **arcsin**, **arccsc**, and **arctan** functions, if we have a negative argument, we’ll end up in **Quadrant IV** (specifically $ \displaystyle -\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$), and for the **arccos**, **arcsec**, and **arccot** functions, if we have a negative argument, we’ll end up in **Quadrant II** ($ \displaystyle \frac{\pi }{2}\le \theta \le \pi $). For arguments outside the domains of the trig functions for **arcsin**, **arccsc**, **arccos**, and **arcsec**, we’ll get **no solution**.

Evaluate the following inverse functions:

## Trig Inverses in the Calculator

To put trig inverses in the graphing calculator, use the **2**^{nd} button before the trig functions like this: ; however, with radians, you won’t get the exact answers with $ \pi $ in it. (In the **degrees** mode, you will get the degrees.) Don’t forget to change to the appropriate mode (radians or degrees) using **DRG** on a TI scientific calculator, or **mode** on a TI graphing calculator.

Here’s an example in **radian mode**: , and in **degree mode**: .

For the **reciprocal functions** (**csc**, **sec**, and **cot**), take the reciprocal of what’s in parentheses, and then use the “normal” trig functions in the calculator. For example, to put $ {{\sec }^{-1}}\left( -\sqrt{2} \right)$ in the calculator (degrees mode), use $ {{\cos }^{-1}}$ as follows: .

When finding the **arccot** or $ {{\cot }^{-1}}$ of a **negative number**, add $ \pi $ to the answer (or **180°** if in degrees); this is because **arccot** comes from **Quadrants I** and **II**, and since we’re using the **arctan** function in the calculator, we need to add $ \pi $. Here is example of getting $ \displaystyle {{\cot }^{-1}}\left( -\frac{1}{\sqrt{3}} \right)$ in radians: , or in degrees: .

**IMPORTANT NOTE**: When getting trig inverses in the calculator, we only get one value back (which we should, because of the **domain restrictions**, and thus **quadrant restrictions**). When **solving trig equations**, however, we typically get many solutions, for example, if we want values in the interval $ \left[ {0,2\pi } \right)$, or over the reals. We’ll see how to use the inverse trig function in the calculator when **solving trig equations** here in the **Solving Trigonometric Equations** section.

## Transformations of the Inverse Trig Functions

We learned how to transform Basic Parent Functions here in the **Parent Functions and Transformations** section, and we learned how to transform the **six Trigonometric Functions**** here. **Now we will transform the **Inverse Trig Functions**.

### T-Charts for the Six Inverse Trigonometric Functions

Some prefer to do all the transformations **with** **t-charts** like we did earlier, and some prefer it **without t-charts**; most of the examples will show t-charts.

Here are the **inverse** **trig parent function t-charts** I like to use. Note that each is in the **correct quadrants** (in order to make true functions). Note that when the original functions have **0**’s as $ y$ values, their respective reciprocal functions are **undefined** (undef) at those points (because of division of **0**); these are **vertical asymptotes**. And remember that **arcsin** and $ {{\sin }^{-1}}$ , for example, are the same thing.

Here are examples, using * t*-charts to perform the transformations. Remember that when functions are transformed on the

**outside**of the function, or parentheses, you move the function up and down and do the “

**regular**” math, and when transformations are made on the

**inside**of the function, or parentheses, you move the function

**back and forth**, but do the “opposite math” (the $ x$-points turn into $ \displaystyle \frac{1}{b}x+c$, and the $ y$-points turn into $ ay+d$ for equations in the form $ y=a{{\sin }^{{-1}}}\left( {b\left( {x-c} \right)} \right)+d$):

Here are examples of reciprocal trig function transformations:

You might also have to come up with an inverse trig equation, given a graph (note that other answers may also be correct):

## Composite Inverse Trig Functions with Special Values/Angles

Sometimes you’ll have to take the trig function of an inverse trig function; sort of “undoing” what you’ve just done (called **composite inverse trig functions**). We still have to remember which quadrants the inverse (inside) trig functions come from: .

Let’s start with some examples with the **special values **or** special angles**, meaning the “answers” will be on the unit circle. Note that the last two aren’t special angles:

### Trig Composites on the Calculator

You can also put trig composites in the graphing calculator (and they don’t have to be special angles), but remember to add $ \pi $ to the answer that you get (or **180°** if in degrees) when you are getting the **arccot** or $ {{\cot }^{{-1}}}$ of a **negative number** (see last example).

Note again for the reciprocal functions, you put **1** over the whole trig function when you work with the regular trig functions (like **cos**), and you take the reciprocal of what’s in the parentheses when you work with the inverse trig functions (like **arccos**).

**Note**: We do have to be careful when using $ \displaystyle \frac{1}{{\tan \left( x \right)}}$ for $ \cot \left( x \right)$ in the calculator. For angles $ \displaystyle \frac{\pi }{2},\frac{{3\pi }}{2}$, the results won’t be correct; it shows an error, instead of **0** (try it!). It would be better to use $ \displaystyle \frac{{\cos \left( x \right)}}{{\sin \left( x \right)}}$ in this case.

## Composite Inverse Trig Functions with Non-Special Angles

You might also have to find composite inverse trig functions with **non-special angles**, which means that they are not found on the Unit Circle. (Examples of **special angles** are **0°, 45°, 60°, 270°,** and their radian equivalents.) The easiest way to do this is to draw triangles on they coordinate system, and (if necessary) use the **Pythagorean Theorem** to find the missing sides.

To know where to put the triangles, use the “**bowtie**” hint: always make the triangle you draw as part of a bowtie that sits on the $ x$-axis. Note that the triangle needs to “hug” the $ x$-axis, not the $ y$-axis:

Find the values of the composite trig functions (inside) by drawing triangles, using **SOH-CAH-TOA**, or the **trig definitions** found here in the **Right Triangle Trigonometry** Section, and then use the **Pythagorean Theorem **to determine the unknown sides. Use **SOH-CAH-TOA** ($ \displaystyle \text{Sin}=\frac{{\text{Opp}}}{{\text{Hyp}}};\,\,\text{Cos}=\frac{{\text{Adj}}}{{\text{Hyp}}};\,\,\text{Tan}=\frac{{\text{Opp}}}{{\text{Adj}}}$) again to find the (outside) trig values.

**Note**: If the angle we’re dealing with is on one of the axes, such as with the arctan(**0°**), we don’t have to draw a triangle, but just draw a line on the $ x$-** **or

**$ y$-axis. Be careful with division by**

**0**; an answer will be “undefined” in these cases.

Here are some problems. Remember again that $ r$ **(hypotenuse of triangle) is never negative**, and when you see whole numbers as arguments, use **1** as the denominator for the triangle. Watch for **negative values**, depending on which quadrant the triangle lies, noting that you’ll **never be drawing a triangle in Quadrant III**. You can check most of these answers in the calculator.

Here are some problems where we have **variables** in the side measurements. Note that the algebraic expressions are still based on the Pythagorean Theorem for the triangles, and that $ r$ (hypotenuse) is never negative.

Assume that all variables are positive, and note that I used the variable $ t$ instead of $ x$ to avoid confusion with the $ x$’s in the triangle. Decide on which quadrant to use by the signs of the variables.

Here are other types of Inverse Trig problems you may see:

**Learn these rules, and practice, practice, practice!**

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the **Mathway** site, where you can register for the **full version** (steps included) of the software. You can even get math worksheets.

You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to **Solving Trigonometric Equations **– you are ready!