This section covers:

**Introduction to Inverse Trig Functions****Graphs of Inverse Functions****Evaluating Inverse Trig Functions – Special Angles****Transformations of the Inverse Trig Functions****Composite Inverse Trig Functions with Non-Special Angles****More Practice**

# 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

**values, and the inverse of a function is symmetrical (a mirror image) around the line**

*y***.**

*\(y=x\)*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**. So, as shown below, we will 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**.

So 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

**above) is the angle (usually in radians).**

*y*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 \(\sin \left( y \right)=x\).

Let’s show how quadrants are important when getting the inverse of a trig function using the **sin** function. In order to make an inverse trig function an actual function, we’ll only take the values between \(-\frac{\pi }{2}\) and \(\frac{\pi }{2}\), so the sin function passes the horizontal line test (meaning its inverse is a function):

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

So the inverse **cos**, **sec**, and **cot** functions will return values in the **I and II Quadrants**, and the inverse **sin**,** csc**,** and tan** functions will return values in the **I and IV Quadrants **(but remember that you need the **negative values** in** Quadrant IV**). (I would just memorize these, since it’s simple to do so). 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 \({{\sin }^{{-1}}}\left( {\frac{{\sqrt{2}}}{2}} \right)\) for example, we only pick the answers from **Quadrants** **I** and **IV**, so we get \(\frac{\pi }{4}\) only. But if we are solving \(\sin \left( x \right)=\frac{{\sqrt{2}}}{2}\) like in the **Solving Trigonometric Functions** section, we get \(\frac{\pi }{4}\) and \(\frac{{3\pi }}{4}\) in the interval (0, 2*π*); there are no domain restrictions.

# Graphs of Inverse Trig Functions

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

Here are the **inverse tan** and **cot** functions. Notice that the **tan** and **cot** inverse functions come from different sets of quadrants: **tan** from Quadrants I and IV, and **cot** from Quadrants I and II:

And here are the **inverse csc** and **sec** functions:

# Evaluating Inverse Trig Functions – Special Angles

When you are asked to evaluate inverse functions, you may be see the notation like \({{\sin }^{-1}}\) or arcsin.

The following examples makes use of the fact that the angles we are evaluating are **special values or special angles**, or angles that have trig values that we can compute exactly (they come right off the **Unit Circle** that we have studied).

Here is the **Unit Circle** again:

To do these problems, use the **Unit Circle** 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 \({{\sin }^{-1}}\left( -\frac{1}{2} \right)\) or \(\arcsin \left( -\frac{1}{2} \right)\), we see that the angle is 330°, or \(\frac{11\pi }{6}\). But since our answer has to be between \(-\frac{\pi }{2}\) and \(\frac{\pi }{2}\), we need to change this to the co-terminal angle \(-30{}^\circ \), or \(-\frac{\pi }{3}\).

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

## Trig Inverses in the Calculator

You can also put trig inverses in the graphing calculator and use the 2^{nd} button before the trig functions: ; however, with radians, you won’t get the exact answers with * π* in it. (In the

**degrees**mode, you will get the degrees.) Here’s an example in

**radian mode**: , and in

**degree mode**: .

For the **reciprocal functions** (**csc**, **sec**, and **cot**), you 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), you’ll use \({{\cos }^{-1}}\) as follows: . Now, when you are getting the **arccot** or \({{\cot }^{-1}}\) of a **negative number**, you have to add * π* to the answer that you get (or 180° if in degrees); this is because

**arccot**come from Quadrants I and II, and since we’re using the

**arctan**function in the calculator, we need to add

*. Here is example of getting \({{\cot }^{-1}}\left( -\frac{1}{\sqrt{3}} \right)\) in radians: , or in degrees: .*

**π****Note**: For all inverse trig functions of a positive argument (other than 1), we should get an angle in Quadrant I (\(0\le \theta \le \frac{\pi }{2}\)). For the arcsin, arccsc, and arctan functions, if we have a negative argument, we’ll end up in Quadrant IV (specifically \(-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}\)), and for the arccos, arcsec, and arccot functions, if we have a negative argument, we’ll end up in Quadrant II (\(\frac{\pi }{2}\le \theta \le \pi \)).

Here are more problems:

# 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 also 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”:

Here are examples of reciprocal trig function transformations:

# 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:

## 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 ** π** 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). (I checked answers for the exact angle solutions).

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**).

Some examples:

# Composite Inverse Trig Functions with Non-Special Angles

You will also have to find the 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.

Remember that the ** r** (hypotenuse)

**can never be negative**!

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

**axis, not the**

*x***axis:**

*y*We 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 using the **Pythagorean Theorem **to determine the unknown sides. Then we use SOH-CAH-TOA again to find the (outside) trig values.** **We still have to remember which quadrants the inverse (inside) trig functions come from:

**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

**axis.**

*y*Let’s do 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. Also note that you’ll

**never be drawing a triangle in Quadrant III**for these problems.

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

**to avoid confusion with the**

*x***’s in the triangle:**

*x***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!

excellent worked out of trigonometry

excellent worked out of inverse function

Why do use radians and degrees when the x axis consisting of the real numbers is being wrapped around the unit circle. The domain of sin x is all reals… not degrees…………technically you compromise the whole concept of the Wrapping Function …

Thanks for writing! I’m not sure exactly what you are asking? Could you point to a particular place on the page? Thanks, Lisa

The last example of composite inverse trig functions with variables, the solution should be squareroot(t^2 +16)/t.

Thank you so so much! I can’t believe I didn’t catch this – I must be staring at this stuff too long 😉 PLEASE let me know if you see anything else. Lisa 🙂