Lesson 8.3 : Inverse Trigonometric Functions

New Concept

Inverse trigonometric functions, like $y = \operatorname{sin}^{-1}(x)$, reverse the process: they take a ratio as input and give an angle as output. We'll explore how to find exact values for these functions and use them in composite expressions.

What’s next

This is just the start! Next, you'll tackle practice cards to master evaluating inverse trig functions and their compositions.

Property

On restricted domains, we can define the inverse trigonometric functions.

• The inverse sine function $y = \sin^{-1} x$ means $x = \sin y$. The inverse sine function is sometimes called the arcsine function, and notated $\arcsin x$.

• The inverse cosine function $y = \cos^{-1} x$ means $x = \cos y$. The inverse cosine function is sometimes called the arccosine function, and notated $\arccos x$.

• The inverse tangent function $y = \tan^{-1} x$ means $x = \tan y$. The inverse tangent function is sometimes called the arctangent function, and notated $\arctan x$.

Examples

• Since $\sin(\frac{\pi}{2}) = 1$, we can write the inverse relationship as $\frac{\pi}{2} = \sin^{-1}(1)$.
• Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, the inverse statement is $\frac{\pi}{3} = \cos^{-1}(\frac{1}{2})$.
• Since $\tan(0) = 0$, it follows that $0 = \tan^{-1}(0)$. The function takes the ratio 0 and returns the angle 0 radians.

Explanation

Because trig functions are periodic, they aren't one-to-one. We must restrict their domains to create a function that has a true inverse. An inverse trig function takes a ratio as input and gives you back the corresponding angle.

Property

For angles in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, if $\sin y = x$, then $\sin^{-1} x = y$.

For angles in the interval $[0, \pi]$, if $\cos y = x$, then $\cos^{-1} x = y$.

For angles in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, if $\tan y = x$, then $\tan^{-1} x = y$.

Property

Given a special input value, evaluate an inverse trigonometric function.

1. Find angle $x$ for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.
2. If $x$ is not in the defined range of the inverse, find another angle $y$ that is in the defined range and has the same sine, cosine, or tangent as $x$.

Examples

• To evaluate $\cos^{-1}(\frac{\sqrt{3}}{2})$, we look for an angle in $[0, \pi]$ where the cosine is $\frac{\sqrt{3}}{2}$. The answer is $\frac{\pi}{6}$.
• To evaluate $\sin^{-1}(-\frac{1}{2})$, we need an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. While $\frac{11\pi}{6}$ works, the correct angle in the range is $-\frac{\pi}{6}$.
• To evaluate $\tan^{-1}(-1)$, we look for an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$ where the tangent is $-1$. The answer is $-\frac{\pi}{4}$.

Explanation

Think of this as working backward on the unit circle. You are given a value (a coordinate) and need to find the angle that produces it, but your answer must be within the inverse function's official range, like $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for sine.

Property

Given two sides of a right triangle, find an angle $\theta$.

1. If the adjacent side $a$ and hypotenuse $h$ are known, use $\theta = \cos^{-1}(\frac{a}{h})$.
2. If the opposite side $p$ and hypotenuse $h$ are known, use $\theta = \sin^{-1}(\frac{p}{h})$.
3. If the opposite side $p$ and adjacent side $a$ are known, use $\theta = \tan^{-1}(\frac{p}{a})$.

Examples

• In a right triangle with an opposite side of 5 and a hypotenuse of 13, the angle $\theta$ is $\sin^{-1}(\frac{5}{13}) \approx 0.395$ radians.
• If a right triangle has a side adjacent to angle $\theta$ of length 8 and a hypotenuse of 17, the angle is $\theta = \cos^{-1}(\frac{8}{17}) \approx 1.08$ radians.
• For a right triangle with an opposite side of 15 and an adjacent side of 8, the angle $\theta$ is $\tan^{-1}(\frac{15}{8}) \approx 1.08$ radians.

Explanation

Inverse trig functions are the key to finding unknown angles in right triangles from side lengths. Use SOHCAHTOA to decide which ratio to use, then apply the corresponding inverse function to find the angle measure.

Property

Compositions of a trigonometric function and its inverse:
$\sin(\sin^{-1} x) = x \text{ for } -1 \leq x \leq 1$
$\cos(\cos^{-1} x) = x \text{ for } -1 \leq x \leq 1$
$\tan(\tan^{-1} x) = x \text{ for } -\infty < x < \infty$

$\sin^{-1}(\sin x) = x \text{ only for } -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$
$\cos^{-1}(\cos x) = x \text{ only for } 0 \leq x \leq \pi$
$\tan^{-1}(\tan x) = x \text{ only for } -\frac{\pi}{2} < x < \frac{\pi}{2}$

Examples

• $\cos(\cos^{-1}(0.7)) = 0.7$ because $0.7$ is in the domain $[-1, 1]$.
• $\cos^{-1}(\cos(\frac{\pi}{5})) = \frac{\pi}{5}$ because $\frac{\pi}{5}$ is in the restricted domain $[0, \pi]$.
• $\sin^{-1}(\sin(\frac{3\pi}{4})) \neq \frac{3\pi}{4}$. Since $\sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}$, the expression becomes $\sin^{-1}(\frac{\sqrt{2}}{2})$, which is $\frac{\pi}{4}$.

Property

To evaluate compositions of the form $f(g^{-1}(x))$, where $f$ and $g$ are different trigonometric functions, we can use geometric relationships.

Let $\theta = g^{-1}(x)$. This defines a right triangle where the ratio of two sides is known. Use the Pythagorean theorem to find the third side. Then, calculate the value of $f(\theta)$ using the triangle's side lengths.

Examples

• Find $\cos(\sin^{-1}(\frac{3}{5}))$. Let $\theta = \sin^{-1}(\frac{3}{5})$, so $\sin(\theta) = \frac{3}{5}$. In a 3-4-5 triangle, the adjacent side is 4. Thus, $\cos(\theta) = \frac{4}{5}$.
• Find $\tan(\cos^{-1}(\frac{12}{13}))$. Let $\theta = \cos^{-1}(\frac{12}{13})$, so $\cos(\theta)=\frac{12}{13}$. The opposite side is $\sqrt{13^2 - 12^2} = 5$. Thus, $\tan(\theta) = \frac{5}{12}$.
• Find a simplified expression for $\sin(\tan^{-1}(x))$. Let $\theta = \tan^{-1}(x)$, so $\tan(\theta) = \frac{x}{1}$. The hypotenuse is $\sqrt{x^2+1^2}$. Thus, $\sin(\theta) = \frac{x}{\sqrt{x^2+1}}$.