The Inverse Trigonometric Functions

This section covers:

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 the inverse of a function is symmetrical (a mirror image) around the line \(y=x\).

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. 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. 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.

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

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 \(\displaystyle -\frac{\pi }{2}\) and \(\displaystyle \frac{\pi }{2}\), so the sin function passes the horizontal line test (meaning its inverse is a function):

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

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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 \(\displaystyle {{\sin }^{{-1}}}\left( {\frac{{\sqrt{2}}}{2}} \right)\) 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, we get \(\displaystyle \frac{\pi }{4}\) and \(\displaystyle \frac{{3\pi }}{4}\) in the interval \(\left( {0,2\pi } \right)\); there are no domain restrictions.

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

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

 

  \(x\) \(y\)
   –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 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:

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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 \(\displaystyle {{\sin }^{-1}}\left( -\frac{1}{2} \right)\) or \(\displaystyle \arcsin \left( -\frac{1}{2} \right)\), we see that 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 \(\displaystyle -\frac{\pi }{3}\).

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  \(\displaystyle {{\cos }^{-1}}\left( -\frac{1}{\sqrt{2}} \right)\), which is \(\displaystyle {{\cos }^{-1}}\left( -\frac{\sqrt{2}}{2} \right)\), which is \(\displaystyle \frac{3\pi }{4}\), or 135°.

Here are some problems; use the Unit Circle to get the angles:

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.)

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 \(\displaystyle \frac{{\sqrt{3}}}{2}\) back for cos (reciprocal), between 0 and \(\pi \) ( and 180°)?

 

\(\displaystyle \frac{\pi }{6}\) or  30°

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

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

 

0   or 

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

What angle gives us \(\displaystyle -\frac{3}{{\sqrt{3}}}=-\frac{3}{{\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}=-\sqrt{3}\) back for tan (reciprocal), 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.

 

What angle gives us \(\displaystyle \frac{1}{{-1}}=-1\) back for tan (reciprocal), 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 \(\displaystyle -\frac{1}{{\sqrt{2}}}=-\frac{{\sqrt{2}}}{2}\) 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

You can also put trig inverses in the graphing calculator and use the 2nd button before the trig functions:  ; however, with radians, you won’t get the exact answers with \(\pi \) 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:  .

When you are getting the arccot or \({{\cot }^{-1}}\) of a negative number, you have to add \(\pi \) 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 \(\pi \). Here is example of getting  \(\displaystyle {{\cot }^{-1}}\left( -\frac{1}{\sqrt{3}} \right)\)  in radians:  , or in degrees:  .

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).

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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.

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”:

Trig Inverse 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 \(x\) 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 \(x\) 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 \(x\) 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 \(x\) values).

 

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).

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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:

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 \(\tan \left( {\frac{{5\pi }}{6}} \right)=-\frac{1}{{\sqrt{3}}}\,\,\,\left( {=-\frac{{\sqrt{3}}}{3}} \right)\).

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

 

 

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 \(\cot \left( {-\frac{\pi }{3}} \right)=-\frac{1}{{\sqrt{3}}}\,\,\,\,\left( {=-\frac{{\sqrt{3}}}{3}} \right)\).

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

 

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.

 

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). (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:

Inverse Trig Problem Calculator Steps Inverse Trig Problem Calculator Steps
\(\displaystyle \tan \left( {{{{\sec }}^{{-1}}}\left( {-\frac{2}{{\sqrt{3}}}} \right)} \right)\) \(\displaystyle {{\tan }^{{-1}}}\left( {\cot \left( {\frac{{3\pi }}{4}} \right)} \right)\)
\(\displaystyle \cot \left( {\text{arcsin}\left( {-\frac{{\sqrt{3}}}{2}} \right)} \right)\) \({{\tan }^{{-1}}}\left( {\text{sec}\left( {1.4} \right)} \right)\)

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 \(x\)-axis, not the \(y\)-axis:

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(), we don’t have to draw a triangle, but just draw a line on the \(x\) or \(y\)-axis.

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.

Composite Inverse Answer Composite Inverse Answer

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

 

Use SOH-CAH-TOA or \(\displaystyle \sin \left( \theta \right)=\frac{y}{r}\) 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 \(x=8\). Since we want sec of this angle, we have \(\displaystyle \sec \left( \theta \right)=\frac{r}{x}=\frac{{17}}{8}\).

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

 

Use SOH-CAH-TOA or \(\displaystyle \cot \left( \theta \right)=\frac{x}{y}\) 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 \(r=\sqrt{{26}}\). Since we want sin of this angle, we have \(\displaystyle \sin \left( \theta \right)=\frac{y}{r}=\frac{1}{{\sqrt{{26}}}}=\frac{{\sqrt{{26}}}}{{26}}\).

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

 

Use SOH-CAH-TOA or \(\displaystyle \sec \left( \theta \right)=\frac{r}{x}\) 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 \(y=5\). Since we want cot of this angle, we have \(\displaystyle \cot \left( \theta \right)=\frac{x}{y}=\frac{{-12}}{5}=-\frac{{12}}{5}\).

\(\tan \left( {{{{\sec }}^{{-1}}}\left( 0 \right)} \right)\) Since \({{\sec }^{{-1}}}\left( 0 \right)\) means the same thing as \(\displaystyle {{\cos }^{{-1}}}\left( {\frac{1}{0}} \right)\), this angle is undefined. (We can also see this by knowing that 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 SOH-CAH-TOA or \(\displaystyle \tan \left( \theta \right)=\frac{y}{x}\) 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 \(r=5\). Since we want sin of this angle, we have \(\displaystyle \sin \left( \theta \right)=\frac{y}{r}=\frac{{-3}}{5}=-\frac{3}{5}\).

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

 

Since \(\displaystyle {{\cos }^{{-1}}}\left( 0 \right)=\frac{\pi }{2}\) or 90° (between Quadrants I and II), we just need to 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:

Composite Inverse Answer Composite Inverse Answer

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

 

Use SOH-CAH-TOA or \(\displaystyle \sec \left( \theta \right)=\frac{r}{x}\) 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 \(y=\sqrt{{{{1}^{2}}-{{{\left( {t-1} \right)}}^{2}}}}\). Since we want sin of this angle, we have \(\displaystyle \sin \left( \theta \right)=\frac{y}{r}=\sqrt{{1-{{{\left( {t-1} \right)}}^{2}}}}\).

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

 

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

Then use Pythagorean Theorem \(\displaystyle {{r}^{2}}={{t}^{2}}+{{3}^{2}}\) to see that \(r=\sqrt{{{{t}^{2}}+9}}\). Since we want sin of this angle, we have \(\displaystyle \sin \left( \theta \right)=\frac{y}{r}=\frac{3}{{\sqrt{{{{t}^{2}}+9}}}}\).

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

 

Use SOH-CAH-TOA  or \(\displaystyle \cos \left( \theta \right)=\frac{x}{r}\) 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 \(y=\sqrt{{1-{{t}^{2}}}}\). Since we want csc of this angle, we have \(\displaystyle \csc \left( \theta \right)=\frac{r}{y}=\frac{1}{{\sqrt{{1-{{t}^{2}}}}}}\).

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

 

Use SOH-CAH-TOA or \(\displaystyle \sec \left( \theta \right)=\frac{r}{x}\) 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 \(y=\sqrt{{4{{t}^{2}}-9}}\). Since we want tan of this angle, we have \(\displaystyle \tan \left( \theta \right)=\frac{y}{x}=\frac{{\sqrt{{4{{t}^{2}}-9}}}}{{-\,3}}\).

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

 

Use SOH-CAH-TOA or \(\displaystyle \tan \left( \theta \right)=\frac{y}{x}\) 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 \(r=\sqrt{{4{{t}^{2}}+1}}\).   Since we want sin of this angle, we have \(\displaystyle \sin \left( \theta \right)=\frac{y}{r}=-\frac{{2t}}{{\sqrt{{4{{t}^{2}}+1}}}}\).

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

 

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

 

Then use Pythagorean Theorem \(\displaystyle {{r}^{2}}={{t}^{2}}+{{4}^{2}}\) to see that \(y=\sqrt{{4{{t}^{2}}-9}}\). Since we want sec of this angle, we have \(\displaystyle \sec \left( \theta \right)=\frac{r}{x}=-\frac{{\sqrt{{{{t}^{2}}+16}}}}{t}\).

 

Learn these rules, and practice, practice, practice!


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On to Solving Trigonometric Equations  – you are ready!

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