Periodic Functions

8.3 Inverse Trigonometric Functions

Inverse Trigonometric Functions

Inverse trig functions reverse the work of trig functions, but they can only do that after we restrict the original trig functions to intervals where they are one-to-one. That restriction is the one piece of this topic that tends to trip people up. We're already familiar with the idea using $f(x)=x^2$. This function generally doesn't have a real inverse, but it has a pseudo inverse. What number makes $x^2=4$ true? Which number gets left out and why?

Principal Values and Restricted Ranges

Trig functions repeat, which means they fail the horizontal line test on their full domains. To define inverses, we pick a specific piece of each function that is one-to-one. That restricted piece is what the inverse undoes.

Restricting the domain of sine

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Fact: Principal Ranges

The three most common inverse trig functions use these principal ranges:

\[ \begin{align*} \sin^{-1}(x) &: -\frac{\pi}{2} \le y \le \frac{\pi}{2} \\ \cos^{-1}(x) &: 0 \le y \le \pi \\ \tan^{-1}(x) &: -\frac{\pi}{2} < y < \frac{\pi}{2} \end{align*} \]

These ranges tell you which angle you are allowed to return.

Example: Evaluate exact inverse trig values

Evaluate the following exactly.

  1. $\sin^{-1}(1)$
  2. $\cos^{-1}\left(-\frac{1}{2}\right)$
  3. $\tan^{-1}(-\sqrt{3})$
Show Solution
  1. We want the angle in the range of inverse sine whose sine is 1.

    $$ \sin^{-1}(1)=\frac{\pi}{2} $$

  2. We want the angle in the range $[0,\pi]$ whose cosine is $-\frac{1}{2}$.

    $$ \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3} $$

  3. We want the angle in the range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ whose tangent is $-\sqrt{3}$.

    $$ \tan^{-1}(-\sqrt{3})=-\frac{\pi}{3} $$

Example: Approximate inverse trig values in degrees

Use a calculator in degree mode to approximate the following.

  1. $\sin^{-1}\left(\frac{4}{9}\right)$
  2. $\cos^{-1}\left(-\frac{12}{13}\right)$
  3. $\tan^{-1}(4)$
Show Solution

Since these are not special unit-circle values, we use a calculator.

  1. $\sin^{-1}\left(\frac{4}{9}\right)\approx 26.39^\circ$
  2. $\cos^{-1}\left(-\frac{12}{13}\right)\approx 157.38^\circ$
  3. $\tan^{-1}(4)\approx 75.96^\circ$

Be careful with domain restrictions. For instance, $\cos^{-1}(1.2)$ would not be real because cosine outputs only between $-1$ and $1$.

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Compositions with Inverse Trig Functions

You might expect $f^{-1}(f(x))$ to just cancel to $x$, but that only works when the angle already sits inside the principal range of the inverse function. If it does not, you have to swap it for the equivalent angle that does.

Warning

Do not automatically cancel a trig function with its inverse. Always check whether the inside angle lies in the principal range of the inverse trig function.

Example: Evaluate a composition carefully

Evaluate

$$ \cos^{-1}\left(\cos\left(\frac{5\pi}{4}\right)\right). $$

Show Solution

At first glance, it is tempting to say the answer is just $\frac{5\pi}{4}$. But that is not allowed because inverse cosine only returns angles in the interval $[0,\pi]$.

So we ask: what angle in $[0,\pi]$ has the same cosine as $\frac{5\pi}{4}$?

\[ \begin{align*} \cos\left(\frac{5\pi}{4}\right) &=-\frac{\sqrt{2}}{2} \end{align*} \]

The angle in $[0,\pi]$ with cosine $-\frac{\sqrt2}{2}$ is $\frac{3\pi}{4}$. Therefore,

$$ \cos^{-1}\left(\cos\left(\frac{5\pi}{4}\right)\right)=\frac{3\pi}{4}. $$

Example: Evaluate several compositions

Evaluate the following.

  1. $\sin^{-1}\left(\sin\left(\frac{7\pi}{6}\right)\right)$
  2. $\tan^{-1}\left(\tan\left(-\frac{5\pi}{4}\right)\right)$
  3. $\cos\left(\cos^{-1}\left(-\frac{1}{3}\right)\right)$
Show Solution
  1. First compute the sine value.

    $$ \sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2} $$

    Now inverse sine must return an angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$:

    $$ \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6} $$

    So

    $$ \sin^{-1}\left(\sin\left(\frac{7\pi}{6}\right)\right)=-\frac{\pi}{6} $$

  2. Tangent has period $\pi$, so

    $$ -\frac{5\pi}{4}+\pi=-\frac{\pi}{4} $$

    and $-\frac{\pi}{4}$ lies in the principal range of inverse tangent. So

    $$ \tan^{-1}\left(\tan\left(-\frac{5\pi}{4}\right)\right)=-\frac{\pi}{4} $$

  3. This one simplifies immediately because cosine and inverse cosine are applied in the correct order to a number in the domain of inverse cosine:

    $$ \cos\left(\cos^{-1}\left(-\frac{1}{3}\right)\right)=-\frac{1}{3} $$

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Example: Evaluate reverse compositions

Evaluate the following.

  1. $\sin\left(\sin^{-1}\left(\frac{3}{5}\right)\right)$
  2. $\tan\left(\tan^{-1}(-2)\right)$
  3. $\cos\left(\cos^{-1}\left(-\frac{4}{7}\right)\right)$
Show Solution

These simplify directly because the trig function is outside and the inverse trig function is inside, so each pair cancels on the domain of the inverse function.

  1. $\sin\left(\sin^{-1}\left(\frac{3}{5}\right)\right)=\frac{3}{5}$
  2. $\tan\left(\tan^{-1}(-2)\right)=-2$
  3. $\cos\left(\cos^{-1}\left(-\frac{4}{7}\right)\right)=-\frac{4}{7}$
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Using Inverse Trig in Triangles and Applications

When the sides of a right triangle are known and we need an angle, inverse trig is exactly the tool for the job.

Fact: Choosing the Right Inverse Trig Function

If you know:

  1. opposite and hypotenuse, use inverse sine
  2. adjacent and hypotenuse, use inverse cosine
  3. opposite and adjacent, use inverse tangent
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Example: Find an angle in a right triangle

A 12-foot ladder is leaning against a wall. The base of the ladder is 9 feet away from the wall. What angle does the ladder make with the ground?

Show Solution

The ladder is the hypotenuse, and the distance along the ground is the side adjacent to the angle we want. So cosine is the most direct choice.

\[ \begin{align*} \cos(\theta)&=\frac{9}{12}=\frac{3}{4} \end{align*} \]

Now apply inverse cosine:

\[ \begin{align*} \theta&=\cos^{-1}\left(\frac{3}{4}\right) \end{align*} \]

Using a calculator,

$$ \theta\approx 41.41^\circ $$

So the ladder makes an angle of about $41.41^\circ$ with the ground.

Example: Find a height from an angle of elevation

A person is standing 50 feet away from a building and measures the angle of elevation to the top of the building to be $30^\circ$. How tall is the building?

Show Solution

This time we know the adjacent side and the angle, and we want the opposite side. That is a tangent setup.

\[ \begin{align*} \tan(30^\circ)&=\frac{h}{50} \end{align*} \]

Now solve for $h$.

\[ \begin{align*} h&=50\tan(30^\circ) \\ &=50\left(\frac{1}{\sqrt3}\right) \\ &=\frac{50\sqrt3}{3} \end{align*} \]

So the exact height is

$$ \frac{50\sqrt3}{3}\text{ feet} $$

and as a decimal,

$$ h\approx 28.87\text{ feet}. $$

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