Inverse Trigonometric Functions

You already know how to press the $\sin^{-1}$ button on a calculator to find a missing angle — that is covered in Finding Missing Angles. This page goes deeper: what do $\arcsin$, $\arccos$, and $\arctan$ actually mean as mathematical functions? Why do they only return certain angles? And how do you evaluate expressions like $\sin(\arccos x)$ without a calculator?

The Notation Problem: sin⁻¹ vs. 1/sin

Before anything else, let’s clear up the most common source of confusion in trigonometry.

These are completely different operations with completely different answers. The $-1$ in $\sin^{-1}$ is not an exponent — it is notation for the inverse function, just like $f^{-1}$ in algebra means “the function that undoes $f$.”

When in doubt, use $\arcsin$, $\arccos$, and $\arctan$ — these names are unambiguous.

Why We Need Restricted Domains

Consider this question: what angle has a sine of $0.5$?

You might say $30°$. That is correct. But $150°$ also has a sine of $0.5$. So do $390°$, $510°$, $-210°$, and infinitely many others — because the sine function repeats every $360°$.

A function can only give you one output for each input. So mathematicians chose a specific interval for each trig function — called the restricted domain — where the function hits every output exactly once. Your calculator uses this restricted domain automatically.

The Three Inverse Functions

Inverse Sine: arcsin

$y = \arcsin(x) \iff \sin(y) = x \quad \text{where } -\frac{\pi}{2} \le y \le \frac{\pi}{2}$

Inverse Cosine: arccos

$y = \arccos(x) \iff \cos(y) = x \quad \text{where } 0 \le y \le \pi$

Inverse Tangent: arctan

$y = \arctan(x) \iff \tan(y) = x \quad \text{where } -\frac{\pi}{2} < y < \frac{\pi}{2}$

Note that $\arctan$ accepts any real number as input (unlike $\arcsin$ and $\arccos$, which are limited to $[-1, 1]$). The output approaches $\pm 90°$ but never reaches it.

Exact Values from the Unit Circle

Many inverse trig values can be evaluated exactly using angles you already know from the unit circle and special angles.

For negative inputs: $\arcsin(-x) = -\arcsin(x)$ and $\arctan(-x) = -\arctan(x)$ (these are odd functions), while $\arccos(-x) = \pi - \arccos(x)$.

Compositions: Evaluating sin(arccos x) and Similar

On AP exams and in college courses, you will need to simplify expressions like $\sin(\arccos x)$. The trick is to draw a reference triangle.

The Reference Triangle Method

To evaluate $\sin(\arccos(x))$:

1. Let $\theta = \arccos(x)$, which means $\cos\theta = x$
2. Draw a right triangle where $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{1}$
3. The adjacent side is $x$, the hypotenuse is $1$
4. By the Pythagorean theorem, the opposite side is $\sqrt{1 - x^2}$
5. Therefore $\sin(\arccos(x)) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$

Example: Evaluate $\sin\!\left(\arccos\!\left(\frac{3}{5}\right)\right)$

Let $\theta = \arccos\!\left(\frac{3}{5}\right)$, so $\cos\theta = \frac{3}{5}$.

Draw the triangle: adjacent $= 3$, hypotenuse $= 5$, opposite $= \sqrt{25 - 9} = \sqrt{16} = 4$.

$\sin\!\left(\arccos\!\left(\frac{3}{5}\right)\right) = \frac{4}{5}$

Common Compositions

Using the reference triangle method, here are the key results:

These follow directly from the Pythagorean theorem applied to the reference triangle. You do not need to memorize them — just draw the triangle each time.

A Subtle Trap: arcsin(sin x) ≠ x (Sometimes)

When $x$ is inside the restricted range, composing a trig function with its inverse gives back $x$:

$\arcsin(\sin(x)) = x \quad \text{only if } -\frac{\pi}{2} \le x \le \frac{\pi}{2}$

But if $x$ is outside the restricted range, the inverse function “folds” the value back into its range.

Example: What is $\arcsin\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right)$?

First: $\sin\!\left(\frac{5\pi}{6}\right) = \sin(150°) = 0.5$

Then: $\arcsin(0.5) = \frac{\pi}{6}$ (not $\frac{5\pi}{6}$, because $\frac{5\pi}{6}$ is outside the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$)

Answer: $\arcsin\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right) = \frac{\pi}{6}$

Common Mistakes

1. Confusing $\sin^{-1}(x)$ with $\csc(x) = \frac{1}{\sin x}$. The notation $\sin^{-1}$ means the inverse function, not the reciprocal. If you are unsure, use $\arcsin$ instead.
2. Forgetting the restricted range. $\arcsin$ only returns values in $[-90°, 90°]$. If you expect an angle in another quadrant, the inverse function will give you the reference angle instead.
3. Assuming $\arcsin(\sin(x)) = x$ always. This is only true when $x$ is in the restricted domain. Otherwise, the output is the equivalent angle within the restricted range.
4. Over-complicating compositions. When asked to find $\cos(\arcsin(\frac{3}{5}))$, draw a reference triangle — do not try to combine identities algebraically. The triangle method is faster and less error-prone.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• $\sin^{-1}(x)$ and $\arcsin(x)$ mean the same thing — the inverse function, not $\frac{1}{\sin x}$
• Each inverse trig function has a restricted range that ensures one output per input: $\arcsin$ returns $[-90°, 90°]$, $\arccos$ returns $[0°, 180°]$, $\arctan$ returns $(-90°, 90°)$
• Use the reference triangle method to evaluate compositions like $\sin(\arccos x)$ — draw the triangle, fill in the sides, read off the answer
• $\arcsin(\sin(x)) = x$ only when $x$ is already in the restricted range
• When in doubt about notation, write $\arcsin$, $\arccos$, $\arctan$ instead of the ambiguous $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$

Return to Trigonometry for more topics in this section.