Inverse Trigonometric Functions

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:

Original Trig Function Inverse Function

$ \displaystyle f\left( x \right)=-4\cos (2x)$, domain $ \displaystyle 0\le x\le \frac{\pi }{4}$

 

Since this is a vertical stretch of the cosine function (range $ \left[ {-1,1} \right]$), the range of this function is $ \left[ {-4,4} \right]$. So, for the original function, we have:

 

Domain: $ \displaystyle \left[ {0,\frac{\pi }{4}} \right]$      Range: $ \left[ {-4,4} \right]$

Switch the $ x$ and the $ y$ to get the inverse function:

$ \displaystyle \begin{align}y=-4\cos (2x);\,\,\,x=-4\cos (2y);\,\,\,\,-\frac{x}{4}=\cos (2y)\\\,\,{{\cos }^{{-1}}}\left( {-\frac{x}{4}} \right)=2y;\,\,\,\,\,\,2y={{\cos }^{{-1}}}\left( {-\frac{x}{4}} \right)\\\,\,y=\frac{{{{{\cos }}^{{-1}}}\left( {-\frac{x}{4}} \right)}}{2};\,\,\,\,\,{{f}^{{-1}}}\left( x \right)=\frac{1}{2}{{\cos }^{{-1}}}\left( {-\frac{x}{4}} \right)\end{align}$

Switch the domain of the range of the original to get the domain and range of the inverse function.

Domain: $ \left[ {-4,4} \right]$       Range: $ \displaystyle \left[ {0,\frac{\pi }{4}} \right]$

(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.

Function and T-Chart Graph of Trig Inverse Function Function and T-Chart Graph of Trig Inverse Function

$ \begin{array}{l}y={{\sin }^{{-1}}}\left( x \right)\text{ or}\\y=\arcsin \left( x \right)\end{array}$

 

  $ x$ $ y$
   –1 $ \displaystyle -\frac{\pi }{2}$
     0    0
     1   $ \displaystyle \frac{\pi }{2}$

Domain: $ \left[ {-1,1} \right]$      Range: $ \displaystyle \left[ {-\frac{\pi }{2},\frac{\pi }{2}} \right]$

$ \begin{array}{l}y={{\cos }^{{-1}}}\left( x \right)\text{ or}\\y=\arccos \left( x \right)\end{array}$

 

  $ x$ $ y$
   –1 $ \pi $
     0 $ \displaystyle \frac{\pi }{2}$
     1  0

Domain: $ \left[ {-1,1} \right]$      Range: $ \left[ {0,\pi } \right]$

$ \begin{array}{l}y={{\tan }^{{-1}}}\left( x \right)\text{ or}\\y=\arctan \left( x \right)\end{array}$

 

 $ x$ $ y$
 und $ \displaystyle -\frac{\pi }{2}$
    –1 $ \displaystyle -\frac{\pi }{4}$
     0 0
     1  $ \displaystyle \frac{\pi }{4}$
 und  $ \displaystyle \frac{\pi }{2}$

(und = undefined)

 Domain: $ \left( {-\infty ,\infty } \right)$     Range: $ \displaystyle \left( {-\frac{\pi }{2},\frac{\pi }{2}} \right)$

    Asymptotes: $ \displaystyle y=-\frac{\pi }{2},\,\,\frac{\pi }{2}$

$ \begin{array}{l}y={{\cot }^{{-1}}}\left( x \right)\text{ or}\\y=\text{arccot}\left( x \right)\end{array}$

 

  $ x$ $ y$
   und $ \pi $
  –1 $ \displaystyle \frac{3\pi }{4}$
     0   $ \displaystyle \frac{\pi }{2}$
     1   $ \displaystyle \frac{\pi }{4}$
 und   0

 Domain: $ \left( {-\infty ,\infty } \right)$     Range: $ \left( {0,\pi } \right)$

Asymptotes: $ y=0,\,\pi $

$ \begin{array}{l}y={{\csc }^{{-1}}}\left( x \right)\text{ or}\\y=\text{arccsc}\left( x \right)\end{array}$

 

 $ x$ $ y$
   –1 $ \displaystyle -\frac{\pi }{2}$
 und    0
     1   $ \displaystyle \frac{\pi }{2}$

 Domain: $ \left( {-\infty ,-1} \right]\cup \left[ {1,\infty } \right)$   Range: $ \displaystyle \left[ {-\frac{\pi }{2},0} \right)\cup \left( {0,\frac{\pi }{2}} \right]$

Asymptote: $ y=0$

$ \begin{array}{l}y={{\sec }^{{-1}}}\left( x \right)\text{ or}\\y=\text{arcsec}\left( x \right)\end{array}$

 

  $ x$ $ y$
   –1 $ \pi $
 und $ \displaystyle \frac{\pi }{2}$
     1 0

Domain: $ \left( {-\infty ,-1} \right]\cup \left[ {1,\infty } \right)$   Range: $ \displaystyle \left[ {0,\frac{\pi }{2}} \right)\cup \left( {\frac{\pi }{2},\pi } \right]$

Asymptote: $ \displaystyle y=\frac{\pi }{2}$

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 arcsinarccsc, 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}$), and for the arccosarcsec, and arccot functions, if we have a negative argument, we’ll end up in Quadrant II ($ \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:

Inverse Function Answer Inverse Function Answer
$ \displaystyle {{\cos }^{{-1}}}\left( {\frac{1}{2}} \right)$

What angle gives us $ \displaystyle \frac{1}{2}$ back for cos, between 0 and $ \pi $ ( and 180°)?

 

$ \displaystyle \frac{\pi }{3}$ or  60°

$ \displaystyle \arcsin \left( {\frac{{\sqrt{2}}}{2}} \right)$

 

What angle gives us $ \displaystyle \frac{{\sqrt{2}}}{2}$ back for sin, between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

 

$ \displaystyle \frac{\pi }{4}$ or  45°

 

$ \arctan \left( {-1} \right)$

What angle gives us –1 back for tan, between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

 

$ \displaystyle -\frac{\pi }{4}$ or  –45°

$ \displaystyle \arccos \left( {-\frac{{\sqrt{3}}}{2}} \right)$

What angle gives us $ \displaystyle -\frac{{\sqrt{3}}}{2}$ back for cos, between 0 and $ \pi $ ( and 180°)?

 

$ \displaystyle \frac{{5\pi }}{6}$ or  150°

$ {{\sin }^{{-1}}}\left( 2 \right)$

 

 

What angle gives us 2 back for sin, between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

 

Since the range of $ {{\sin }^{{-1}}}$ (domain of sin) is $ \left[ {-1,1} \right]$, this is undefined, or no solution, or $ \emptyset $.

$ \displaystyle {{\sec }^{{-1}}}\left( {\frac{2}{{\sqrt{3}}}} \right)$

What angle gives us $ \displaystyle \frac{2}{{\sqrt{3}}}$ back for sec (reciprocal $ \displaystyle \frac{{\sqrt{3}}}{2}$ back for cos), between 0 and $ \pi $ ( and 180°)?

$ \displaystyle \frac{\pi }{6}$ or  30°

$ \text{arcsec}\left( 1 \right)$

What angle gives us $ 1$ back for sec (reciprocal $ \displaystyle \frac{1}{1}=1$ back for cos), between and $ \pi $ (0 and 180°)?

 

0   or 

$ \displaystyle \text{arccot}\left( {-\frac{{\sqrt{3}}}{3}} \right)$

What angle gives us $ \displaystyle -\frac{{\sqrt{3}}}{3}$ back for cot (reciprocal $ \displaystyle -\frac{3}{{\sqrt{3}}}=-\frac{3}{{\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}=-\sqrt{3}$ back for tan), between 0 and $ \pi $ ( and 180°)?

$ \displaystyle \frac{{2\pi }}{3}$  or  120°

Note that if we put $ {{\tan }^{{-1}}}\left( {-\sqrt{3}} \right)$ in the calculator, we would have to add $ \pi $ (or 180°) so it will be in Quadrant II.

$ {{\cot }^{{-1}}}\left( {-1} \right)$

What angle gives us $ -1$ back for cot (reciprocal $ \displaystyle \frac{1}{{-1}}=-1$ back for tan), between 0 and $ \pi $ ( and 180°)?

$ \displaystyle \frac{{3\pi }}{4}$ or  135°

Note that if we put $ {{\cot }^{{-1}}}\left( {-1} \right)$ in the calculator, we would have to add $ \pi $ (or 180°) so it will be in Quadrant II.

$ \text{arccsc}\left( {-\sqrt{2}} \right)$

What angle gives us $ \displaystyle -\sqrt{2}$ back for csc (reciprocal $ \displaystyle -\frac{1}{{\sqrt{2}}}=-\frac{{\sqrt{2}}}{2}$ back for sin), between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

 

$ \displaystyle -\frac{\pi }{4}$ or  ­–45°

Trig Inverses in the Calculator

To put trig inverses in the graphing calculator, use the 2nd 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.

Inverse Trig Function T-Charts

$ y={{\sin }^{{-1}}}\left( x \right)$

   x y
   –1 $ \displaystyle -\frac{\pi }{2}$
     0 0
     1 $ \displaystyle \frac{\pi }{2}$

 

$ y={{\cos }^{{-1}}}\left( x \right)$

x
   –1 $ \pi $
     0 $ \displaystyle \frac{\pi }{2}$
     1 0

 

$ y={{\tan }^{{-1}}}\left( x \right)$

x y
  undef $ \displaystyle -\frac{\pi }{2}$
    –1 $ \displaystyle -\frac{\pi }{4}$
      0 0
      1 $ \displaystyle \frac{\pi }{4}$
 undef $ \displaystyle \frac{\pi }{2}$

$ y={{\csc }^{{-1}}}\left( x \right)$

x y
   –1 $ \displaystyle -\frac{\pi }{2}$
  undef 0
     1 $ \displaystyle \frac{\pi }{2}$

 

$ y={{\sec }^{{-1}}}\left( x \right)$

 x y
   –1 $ \pi $
undef $ \displaystyle \frac{\pi }{2}$
     1 0

 

$ y={{\cot }^{{-1}}}\left( x \right)$

 x y
 undef $ \pi $
    –1 $ \displaystyle \frac{3\pi }{4}$
      0 $ \displaystyle \frac{\pi }{2}$
      1 $ \displaystyle \frac{\pi }{4}$
  undef 0

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$):

Inverse Trig Transformation T-Chart Graph

$ \displaystyle y={{\sin }^{{-1}}}\left( {2x} \right)-\frac{\pi }{2}$

 

 

Graph is stretched horizontally by factor of $ \displaystyle \frac{1}{2}$ (compression).

Graph is moved down $ \displaystyle \frac{\pi }{2}$ units.

$ \frac{1}{2}x$    $ x$ $ y$    $ y-\frac{\pi }{2}$
$ \displaystyle -\frac{1}{2}$     -1 $ \displaystyle -\frac{\pi }{2}$    $ -\pi $
  0       0  0     $ \displaystyle -\frac{\pi }{2}$
$ \displaystyle \frac{1}{2}$     1 $ \displaystyle \frac{\pi }{2}$       0

 

Domain: $ \displaystyle \left[ {-\frac{1}{2},\frac{1}{2}} \right]$

Range: $ \left[ {-\pi,0} \right]$

$ \displaystyle y=4{{\cos }^{{-1}}}\left( {\frac{x}{2}} \right)$

 

 

Graph is stretched vertically by a factor of 4.

 

Graph is stretched horizontally by factor of 2.

$ 2x$    $ x$ $ y$     $ 4y$
–2     -1 $ \pi $    $ 4\pi $
 0      0 $ \displaystyle \frac{\pi }{2}$     $ 2\pi $
  2       1 0       0

 

Domain: $ \left[ {-2,2} \right]$

Range: $ \left[ {0,4\pi } \right]$

$ y=-3\arctan \left( {x+1} \right)$

 

 

Graph is flipped over the $ x$-axis and stretched by a factor of 3.

 

Graph is shifted to the left 1 unit.

$ x-1$  $ x$ $ y$  $ -3y$
und  und $ \displaystyle -\frac{\pi }{2}$      $ \displaystyle \frac{{3\pi }}{2}$
–2   –1 $ \displaystyle -\frac{\pi }{4}$      $ \displaystyle \frac{{3\pi }}{4}$
–1    0 0      0
  0    1 $ \displaystyle \frac{\pi }{4}$    $ \displaystyle -\frac{3\pi }{4}$
und  und $ \displaystyle \frac{\pi }{2}$  $ \displaystyle -\frac{3\pi }{2}$

(und = undefined)

 

Domain: $ \left( {-\infty ,\infty } \right)$

Range:$ \displaystyle \left( {-\frac{{3\pi }}{2}\,,\frac{{3\pi }}{2}\,} \right)$

Asymptotes: $ \displaystyle y=-\frac{{3\pi }}{2},\,\,\frac{{3\pi }}{2}$

(Transform asymptotes as you would the $ y$-values).

Here are examples of reciprocal trig function transformations:

Transformation T-Chart Graph

$ \displaystyle y=-{{\sec }^{{-1}}}\left( {\frac{x}{3}} \right)-\frac{\pi }{2}$

 

 

Graph is flipped over the $ x$-axis and stretched horizontally by factor of 3.

 

Graph is moved down $ \displaystyle \frac{\pi }{2}$ units.

$ 3x$    $ x$ $ y$   $ -y-\frac{\pi }{2}$
 –3     -1 $ \displaystyle \pi $     $ \displaystyle -\frac{{3\pi }}{2}$
 und  und $ \displaystyle \frac{{\pi }}{2}$      $ -\pi $
   3      1 0     $ \displaystyle -\frac{\pi }{2}$

 

Domain: $ \left( {-\infty ,-3} \right]\cup \left[ {3,\infty } \right)$

Range:$ \displaystyle \left[ {-\frac{{3\pi }}{2},-\pi } \right)\cup \left( {-\pi ,\,\,\frac{{3\pi }}{2}} \right]$

Asymptote: $ y=-\pi $

(Transform asymptotes as you would $ y$-values).

$ \displaystyle y=4{{\cot }^{{-1}}}\left( x \right)+\frac{\pi }{4}$

 

 

Graph is stretched vertically by factor of 4.

 

Graph is moved up $ \displaystyle \frac{\pi }{4}$ units.

 

$ x$ $ y$   $ 4y+\frac{\pi }{4}$
  und $ \pi $     $ \displaystyle \frac{{17\pi }}{4}$
–1 $ \displaystyle \frac{{3\pi }}{4}$     $ \displaystyle \frac{{13\pi }}{4}$
 0 $ \displaystyle \frac{{\pi }}{2}$     $ \displaystyle \frac{{9\pi }}{4}$
 1 $ \displaystyle \frac{{\pi }}{4}$     $ \displaystyle \frac{{5\pi }}{4}$
und 0       $ \displaystyle \frac{{\pi }}{4}$

 

Domain: $ \left( {-\infty ,\infty } \right)$

Range: $ \displaystyle \left( {\frac{\pi }{4}\,,\frac{{17\pi }}{4}\,} \right)$

Asymptotes: $ \displaystyle y=\frac{\pi }{4},\,\,\frac{{17\pi }}{4}$

(Transform asymptotes as you would $ y$-values).

$ \begin{array}{l}y=\text{arccsc}\left( {2x-4} \right)-\pi \\y=\text{arccsc}\left( {2\left( {x-2} \right)} \right)-\pi \end{array}$

 

(Factor first to get $ x$ by itself in the parentheses.)

 

 

Graph is stretched horizontally by a factor of $ \displaystyle \frac{1}{2}$ (compression).

 

Graph is shifted to the right 2 units and down $ \pi $ units.

$ \frac{1}{2}x+2$  $ x$ $ y$     $ y-\pi $
$ \displaystyle \frac{3}{2}$       -1 $ \displaystyle -\frac{\pi }{2}$   $ \displaystyle -\frac{3\pi }{2}$
und   und 0    $ -\pi $
$ \displaystyle \frac{5}{2}$       1 $ \displaystyle \frac{\pi }{2}$    $ \displaystyle -\frac{\pi }{2}$

 

Domain: $ \displaystyle \left( {-\infty ,\frac{3}{2}} \right]\cup \left[ {\frac{5}{2},\infty } \right)$

Range: $ \displaystyle \left[ {-\frac{{3\pi }}{2},-\pi } \right)\cup \left( {-\pi ,-\frac{\pi }{2}} \right]$

Asymptote: $ y=-\pi $

(Transform asymptotes as you would $ y$-values).

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

Inverse Trig Graph Equation
What is a possible equation for the following graph?

The shape of the graph is a $ {{\sin }^{{-1}}}$ function; start with $ y=a{{\sin }^{{-1}}}\left( {b\left( {x-c} \right)} \right)+d$. The graph appears to be shifted up $ \pi $ units and reflected vertically, but not shifted horizontally. Now we have $ y=-a{{\sin }^{{-1}}}\left( x \right)+\pi $.

 

The graph appears to be stretched vertically. Normally, the  graph goes up from 0 up to $ \displaystyle \frac{\pi }{2}$ and down to $ \displaystyle -\frac{\pi }{2}$, but this graph goes up three times as much (from $ \pi $ to $ \displaystyle \frac{{5\pi }}{2}$, for example); thus, the vertical stretch is 3. The graph could be $ y=-3{{\sin }^{{-1}}}\left( x \right)+\pi $. Tricky!

What is a possible equation for the following graph?

The shape of the graph is a $ {{\tan }^{{-1}}}$ or $ {{\cot }^{{-1}}}$ function; start with $ y=a{{\tan }^{{-1}}}\left( {b\left( {x-c} \right)} \right)+d$. The graph appears to be shifted left $ 1$ units and reflected vertically. Now we have $ y=-a{{\tan }^{{-1}}}\left( {x+1} \right)$.

 

The graph appears to be stretched vertically. Normally, the $ {{\tan }^{{-1}}}$ graph goes up from $ \displaystyle -\frac{\pi }{2}$ to $ \displaystyle \frac{\pi }{2}$, but this graph goes up from $ -\pi $ to $ \displaystyle \pi $; thus, the vertical stretch is 2. The graph could be $ y=-2{{\tan }^{{-1}}}\left( {x+1} \right)$.

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:

 

Composite Inverse Answer Composite Inverse Answer
 

$ \displaystyle \tan \left( {{{{\sec }}^{{-1}}}\left( {-\frac{2}{{\sqrt{3}}}} \right)} \right)$

 

What angle gives us $ \displaystyle -\frac{2}{{\sqrt{3}}}$ back for sec ($ \displaystyle -\frac{{\sqrt{3}}}{2}$ for cos), between 0 and $ \pi $ ( and 180°)?

$ \displaystyle \frac{{5\pi }}{6}$ or  150°

Since we want tan of this angle, we have $ \displaystyle \tan \left( {\frac{{5\pi }}{6}} \right)=-\frac{1}{{\sqrt{3}}}\,\,\,\left( {=-\frac{{\sqrt{3}}}{3}} \right)$.

$ \cot \left( {\text{arctan}\left( {-\sqrt{3}} \right)} \right)$

 

 

What angle gives us $ -\sqrt{3}$ back for tan, between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

$ \displaystyle -\frac{\pi }{3}$ or  –60°

Since we want cot of this angle, we have $ \displaystyle \cot \left( {-\frac{\pi }{3}} \right)=-\frac{1}{{\sqrt{3}}}\,\,\,\,\left( {=-\frac{{\sqrt{3}}}{3}} \right)$.

$ \displaystyle \tan \left( {{{{\cos }}^{{-1}}}\left( {-\frac{1}{2}} \right)} \right)$

 

What angle gives us $ \displaystyle -\frac{1}{2}$ back for cos, between 0 and $ \pi $ ( and 180°)?

 

$ \displaystyle \frac{{2\pi }}{3}$ or  120°

 

Since we want tan of this angle, we have $ \displaystyle \tan \left( {\frac{{2\pi }}{3}} \right)=-\sqrt{3}$.

$ \cos \left( {{{{\cos }}^{{-1}}}\left( 2 \right)} \right)$ Note that $ {{\cos }^{{-1}}}\left( 2 \right)$ is undefined, since the range of cos (domain of $ {{\cos }^{{-1}}}$) is $ [–1,1]$. Since this angle is undefined, the cos back of this angle is undefined (or no solution, or $ \emptyset $).

 

Notice that just “undoing” an angle doesn’t always work: the answer is not 2.

$ \displaystyle {{\sin }^{{-1}}}\left( {\sin \left( {\frac{{2\pi }}{3}} \right)} \right)$

 

Since $ \displaystyle \sin \left( {\frac{{2\pi }}{3}} \right)=\frac{{\sqrt{3}}}{2}$, what angle that gives us $ \displaystyle \frac{{\sqrt{3}}}{2}$ back for sin, between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

 

$ \displaystyle \frac{\pi }{3}$ or  60°

 

Notice that just “undoing” an angle doesn’t always work: the answer is not $ \displaystyle \frac{{2\pi }}{3}$ (in Quadrant II), but $ \displaystyle \frac{\pi }{3}$ (Quadrant I).

$ \displaystyle \arcsin \left( {\cos \left( {\frac{{3\pi }}{4}} \right)} \right)$

 

Since $ \displaystyle \cos \left( {\frac{{3\pi }}{4}} \right)=-\frac{{\sqrt{2}}}{2}$, what angle that gives us $ \displaystyle -\frac{{\sqrt{2}}}{2}$ back for sin, between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°)?

 

$ \displaystyle -\frac{\pi }{4}$ or  –45°

 

Note again the change in quadrants of the angle.

$ \displaystyle {{\sin }^{{-1}}}\left( {\sin \left( {\frac{{7\pi }}{5}} \right)} \right)$

 

 

This is tricky, since the angle given isn’t a special angle (one on the Unit Circle). But we can look at the Unit Circle to figure it out!

 

First determine the quadrant angle $ \displaystyle \frac{{7\pi }}{5}$; this is Quadrant III. We need an angle back between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°) with the same sin value; this will be in Quadrant IV. Since the reference angle from the $ x$-axis is the same to $ \displaystyle \frac{{7\pi }}{5}$ to $ \displaystyle -\frac{{2\pi }}{5}$, the answer is $ \displaystyle -\frac{{2\pi }}{5}$.

$ \displaystyle {{\tan }^{{-1}}}\left( {\tan \left( {\frac{{7\pi }}{{10}}} \right)} \right)$

 

 

This is tricky, since the angle given isn’t a special angle (one on the Unit Circle). But we can look at the Unit Circle to figure it out!

 

First determine the quadrant angle $ \displaystyle \frac{{7\pi }}{{10}}$; this is Quadrant II. We need an angle back between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (–90° and 90°) with the same tan value; this will be in Quadrant IV. Since the period of tan is $ \pi $, we can just add this to $ \displaystyle \frac{{7\pi }}{{10}}$ to get an angle in Quadrant IV: $ \displaystyle \frac{{7\pi }}{{10}}+\frac{{10\pi }}{{10}}=\frac{{17\pi }}{{10}}$, which is the same angle as $ \displaystyle -\frac{{3\pi }}{{10}}$ .

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.

Here are some examples:

Inverse Trig Problem Calculator Steps Inverse Trig Problem Calculator Steps
$ \displaystyle \tan \left( {{{{\sec }}^{{-1}}}\left( {-\frac{2}{{\sqrt{3}}}} \right)} \right)$ $ \displaystyle {{\tan }^{{-1}}}\left( {\cot \left( {\frac{{3\pi }}{4}} \right)} \right)$
$ \displaystyle \cot \left( {\text{arcsin}\left( {-\frac{{\sqrt{3}}}{2}} \right)} \right)$ $ \displaystyle {{\cot }^{{-1}}}\left( {\text{tan}\left( {\frac{{7\pi }}{4}} \right)} \right)$

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(), 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.

Composite Inverse Answer Composite Inverse Answer

$ \displaystyle \sec \left( {{{{\sin }}^{{-1}}}\left( {\frac{{15}}{{17}}} \right)} \right)$

 

Use $ \displaystyle \sin \left( \theta \right)=\frac{y}{r}$ to see that $ y=15$ and $ r=17$ (Quadrant I).

 

Then use Pythagorean Theorem $ \left( {{{x}^{2}}+{{{15}}^{2}}={{{17}}^{2}}} \right)$ to see that $ x=8$. The sec of this angle is $ \displaystyle \sec \left( \theta \right)=\frac{r}{x}=\frac{{17}}{8}$.

$ \sin \left( {\text{arccot}\left( 5 \right)} \right)$

 

Use $ \displaystyle \cot \left( \theta \right)=\frac{x}{y}$ to see that $ x=5$ and $ y=1$ (for whole numbers, denominator is 1) (Quadrant I).

 

Then use Pythagorean Theorem $ \left( {{{1}^{2}}+{{5}^{2}}={{r}^{2}}} \right)$ to see that $ r=\sqrt{{26}}$. The sin of this angle is $ \displaystyle \sin \left( \theta \right)=\frac{y}{r}=\frac{1}{{\sqrt{{26}}}}=\frac{{\sqrt{{26}}}}{{26}}$.

$ \displaystyle \cot \left( {\text{arcsec} \left( {-\frac{{13}}{{12}}} \right)} \right)$

 

Use $ \displaystyle \sec \left( \theta \right)=\frac{r}{x}$ to see that $ r=13$ and $ x=-12$ (Quadrant II).

 

Then use Pythagorean Theorem $ \left( {{{{\left( {-12} \right)}}^{2}}+{{y}^{2}}={{{13}}^{2}}} \right)$ to see that $ y=5$. The cot of this angle, is $ \displaystyle \cot \left( \theta \right)=\frac{x}{y}=\frac{{-12}}{5}=-\frac{{12}}{5}$.

$ \tan \left( {{{{\sec }}^{{-1}}}\left( 0 \right)} \right)$ The $ {{\sec }^{{-1}}}\left( 0 \right)$ is undefined, since the domain of $ {{\sec }^{{-1}}}$ does not include 0.

 

Since this angle is undefined, the tan of this angle is undefined (or no solution, or $ \emptyset $).

$ \displaystyle \sin \left( {{{{\tan }}^{{-1}}}\left( {-\frac{3}{4}} \right)} \right)$

 

Use $ \displaystyle \tan \left( \theta \right)=\frac{y}{x}$ to see that $ y=-3$ and $ x=4$ (use Quadrant IV).

 

Then use Pythagorean Theorem $ \left( {{{{\left( {-3} \right)}}^{2}}+{{4}^{2}}={{5}^{2}}} \right)$ to see that $ r=5$. The sin of this angle is $ \displaystyle \sin \left( \theta \right)=\frac{y}{r}=\frac{{-3}}{5}=-\frac{3}{5}$.

$ \sin \left( {{{{\cos }}^{{-1}}}\left( 0 \right)} \right)$

 

Since $ \displaystyle {{\cos }^{{-1}}}\left( 0 \right)=\frac{\pi }{2}$ or 90° (between Quadrants I and II), take the $ \displaystyle \sin \left( {\frac{\pi }{2}} \right)$, which is 1.

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.

Composite Inverse Answer Composite Inverse Answer

$ \displaystyle \sin \left( {{{{\sec }}^{{-1}}}\left( {\frac{1}{{t-1}}} \right)} \right)$

 

Use $ \displaystyle \sec \left( \theta \right)=\frac{r}{x}$ to see that $ r=1$ and $ x=t-1$ (Quadrant I).

 

Then use Pythagorean Theorem $ \displaystyle {{y}^{2}}={{1}^{2}}-{{\left( {t-1} \right)}^{2}}$ to see that $ y=\sqrt{{{{1}^{2}}-{{{\left( {t-1} \right)}}^{2}}}}$. Thus, $ \displaystyle \sin \left( \theta \right)=\frac{y}{r}=\sqrt{{1-{{{\left( {t-1} \right)}}^{2}}}}$.

$ \displaystyle \sin \left( {\text{arccot}\left( {\frac{t}{3}} \right)} \right)$

 

Use $ \displaystyle \cot \left( \theta \right)=\frac{x}{y}$ to see that $ x=t$ and $ y=3$ (Quadrant I).

 

Then use Pythagorean Theorem $ \displaystyle {{r}^{2}}={{t}^{2}}+{{3}^{2}}$ to see that $ r=\sqrt{{{{t}^{2}}+9}}$. Thus, $ \displaystyle \sin \left( \theta \right)=\frac{y}{r}=\frac{3}{{\sqrt{{{{t}^{2}}+9}}}}$.

$ \csc \left( {{{{\cos }}^{{-1}}}\left( {-t} \right)} \right)$

 

Use $ \displaystyle \cos \left( \theta \right)=\frac{x}{r}$ to see that $ x=-t$ and $ r=1$ (Quadrant II).

 

Then use Pythagorean Theorem $ \displaystyle {{y}^{2}}={{1}^{2}}-{{\left( {-t} \right)}^{2}}$ to see that $ y=\sqrt{{1-{{t}^{2}}}}$. Thus, $ \displaystyle \csc \left( \theta \right)=\frac{r}{y}=\frac{1}{{\sqrt{{1-{{t}^{2}}}}}}$.

$ \displaystyle \tan \left( {\text{arcsec}\left( {-\frac{2}{3}t} \right)} \right)$

 

Use $ \displaystyle \sec \left( \theta \right)=\frac{r}{x}$ to see that $ r=2t$ and $ x=-3$ (Quadrant II).

 

Then use Pythagorean Theorem $ \displaystyle {{y}^{2}}={{\left( {2t} \right)}^{2}}-{{\left( {-3} \right)}^{2}}$ to see that $ y=\sqrt{{4{{t}^{2}}-9}}$. Thus, $ \displaystyle \tan \left( \theta \right)=\frac{y}{x}=\frac{{\sqrt{{4{{t}^{2}}-9}}}}{{-\,3}}=-\frac{{\sqrt{{4{{t}^{2}}-9}}}}{3}$.

$ \sin \left( {{{{\tan }}^{{-1}}}\left( {-2t} \right)} \right)$

 

Use $ \displaystyle \tan \left( \theta \right)=\frac{y}{x}$ to see that $ y=-2t$ and $ x=1$ (Quadrant IV).

 

Then use Pythagorean Theorem $ \displaystyle {{r}^{2}}={{\left( {-2t} \right)}^{2}}+{{1}^{2}}$ to see that $ r=\sqrt{{4{{t}^{2}}+1}}$. Thus, $ \displaystyle \sin \left( \theta \right)=\frac{y}{r}=-\frac{{2t}}{{\sqrt{{4{{t}^{2}}+1}}}}$.

$ \displaystyle \text{sec}\left( {{{{\tan }}^{{-1}}}\left( {\frac{4}{t}} \right)} \right)$

 

Use $ \displaystyle \tan \left( \theta \right)=\frac{y}{x}$ to see that $ y=4$ and $ x=t$ (Quadrant I).

 

Then use Pythagorean Theorem $ \displaystyle {{r}^{2}}={{t}^{2}}+{{4}^{2}}$ to see that $ r=\sqrt{{{{t}^{2}}+16}}$. Thus, $ \displaystyle \sec \left( \theta \right)=\frac{r}{x}=\frac{{\sqrt{{{{t}^{2}}+16}}}}{t}$.

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

 

Inverse Trig Problem Solution
How many solution(s) does $ {{\cos }^{{-1}}}x$ have, if $ x$ is a single value in the interval $ \left[ {-1,1} \right]$? The graph of $ {{\cos }^{{-1}}}$ is:

We see that there is only one solution, or $ y$ value, for each $ x$ value. This makes sense since the function is one-to-one (has to pass the vertical line test).

For what values of $ x$ is $ {{\sin }^{{-1}}}\left( {\sin \left( x \right)} \right)=x$?

 

For what values of $ x$ is $ \sec \left( {{{{\sec }}^{{-1}}}\left( x \right)} \right)=x$?

For problems like these, we need to look at any restrictions on the inverse functions, with respect to the $ x$. For the first problem, the “$ x$” part is the range of $ {{\sin }^{{-1}}}$, so $ x$ is restricted to $ \displaystyle \left[ {-\frac{\pi }{2},\frac{\pi }{2}} \right]$.

In the second problem, the “$ x$” part is the domain of $ {{\sec }^{{-1}}}$, so $ x$ is restricted to $ \left( {-\infty ,-1} \right]\cup \left[ {1,\infty } \right)$.

$ \sin \left( {{{{\sin }}^{{-1}}}\left( x \right)} \right)=x$ is true for which of the following value(s)?

a) $ \displaystyle -\frac{{\sqrt{3}}}{2}$      b)  0       c) $ \displaystyle \frac{1}{{\sqrt{2}}}$      d)  3

All answers are true, except for d), since 3 is not in the domain of $ {{\sin }^{{-1}}}$ ($ \left[ {-1,1} \right]$). Try the other values, using the Unit Circle; they work! You can also put these in the calculator to see if they work.
$ {{\tan }^{{-1}}}\left( {\tan \left( x \right)} \right)=x$ is true for which of the following value(s)?

a) $ \displaystyle \frac{{5\pi }}{3}$       b)  0        c) $ \displaystyle -\frac{\pi }{3}$       d)  3

Only b) and c) work since these values are between $ \displaystyle -\frac{\pi }{2}$ and $ \displaystyle \frac{\pi }{2}$ (the range of $ {{\tan }^{{-1}}}$). Try these using the Unit Circle. You can also put these in the calculator to see if they work.
Which of the following are undefined?

a) $ \displaystyle {{\csc }^{{-1}}}\left( {\frac{{13}}{2}} \right)$  b) $ \displaystyle {{\sin }^{{-1}}}\left( {\frac{4}{{\sqrt{{15}}}}} \right)$  c) $ \displaystyle {{\cot }^{{-1}}}\left( {-\frac{{13}}{2}} \right)$

b) is undefined since the value is outside the domain of the $ {{\sin }^{{-1}}}$ function ($ \left[ {-1,1} \right]$). The others are defined, since they fall in the domains of the inverse trig functions. You can also put these in the calculator to see if they work.
What are the asymptotes of $ y=8{{\cot }^{{-1}}}\left( {4x+1} \right)$? Since the horizontal asymptotes of the parent $ {{\cot }^{{-1}}}$ function are $ 0$ and $ \pi$, replace the $ {{\cot }^{{-1}}}$ function with these values and solve for $ y$ (vertical stretch):

$ \begin{array}{c}y=8\left( 0 \right)\,\,\,\,\,\,\,\,y=8\left( \pi \right)\\y=0\,\,\,\,\,\,\,\,\,y=8\pi \end{array}$

The asymptotes are $ y=0$ and $ y=8\pi $; these are horizontal.

Given $ f\left( x \right)=\sin \left( {{{{\cot }}^{{-1}}}\left( {-.4} \right)} \right)$, which of the following are true?

 

       a) $ \displaystyle f\left( x \right)>0$          b)$ \displaystyle f\left( x \right)=0$

       c) $ \displaystyle f\left( x \right)<0$          d) $ \displaystyle f\left( x \right)$is undefined

a) is correct, since the value of $ {{\cot }^{{-1}}}\left( {-.4} \right)$ will end up in Quadrant II, where $ f\left( x \right)=\sin $ is positive.

 

You can also put this in the calculator, but remember when we take $ {{\cot }^{{-1}}}\left( {\text{negative number}} \right)$, we have to add $ \pi $ to the value we get.

Solve the following inverse trig problem:

 

$ 12{{\sin }^{{-1}}}\left( {3x} \right)=3\pi $

Solve for $ {{\sin }^{{-1}}}$, and then the argument of this (the $ 3x$) is not an angle, but the actual trig value:

$ \begin{align}12{{\sin }^{{-1}}}\left( {3x} \right)=3\pi &;\,\,\,\,{{\sin }^{{-1}}}\left( {3x} \right)=\frac{\pi }{4}\\3x=\frac{{\sqrt{2}}}{2}&;\,\,\,\,x=\frac{{\sqrt{2}}}{6}\end{align}$

Check it in a calculator; it works!

Learn these rules, and practice, practice, practice!


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On to Solving Trigonometric Equations  – you are ready!