Inverse Trigonometric Functions Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Inverse Trigonometric Functions.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Functions that reverse the trigonometric functions: given a ratio, they return the corresponding angle. \arcsin, \arccos, and \arctan are the inverses of \sin, \cos, and \tan on restricted domains.

Regular trig functions answer: 'Given an angle, what's the ratio?' Inverse trig functions answer the reverse: 'Given a ratio, what's the angle?' Since \sin and \cos are many-to-one (many angles give the same ratio), we must restrict their domains to make the inverse a proper function. Think of it like this: if you know the slope of a ramp is 0.5, \arcsin(0.5) = 30° tells you the angle.

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How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Inverse trig functions exist only because we restrict the original trig functions to intervals where they are one-to-one: \sin is restricted to [-\frac{\pi}{2}, \frac{\pi}{2}], \cos to [0, \pi], and \tan to (-\frac{\pi}{2}, \frac{\pi}{2}).

Common stuck point: The restricted domains mean inverse trig functions only return angles in specific ranges. \arcsin returns values in [-\frac{\pi}{2}, \frac{\pi}{2}], so \arcsin(-\frac{1}{2}) = -\frac{\pi}{6}, NOT \frac{7\pi}{6}.

Sense of Study hint: Write down the allowed output range first (e.g., arcsin returns angles in [-pi/2, pi/2]). Then find the angle in that range whose sine matches.

Worked Examples

Example 1

easy
Evaluate \arcsin\!\left(\frac{1}{2}\right), \arccos\!\left(-\frac{\sqrt{2}}{2}\right), and \arctan(1). State the range of each inverse trig function.

Solution

  1. 1
    \arcsin(\frac{1}{2}): find \theta\in[-\pi/2,\pi/2] with \sin\theta=\frac{1}{2}. Answer: \theta=\frac{\pi}{6}.
  2. 2
    \arccos(-\frac{\sqrt{2}}{2}): find \theta\in[0,\pi] with \cos\theta=-\frac{\sqrt{2}}{2}. Answer: \theta=\frac{3\pi}{4} (Q2).
  3. 3
    \arctan(1): find \theta\in(-\pi/2,\pi/2) with \tan\theta=1. Answer: \theta=\frac{\pi}{4}.

Answer

\arcsin(\frac{1}{2})=\frac{\pi}{6}; \arccos(-\frac{\sqrt{2}}{2})=\frac{3\pi}{4}; \arctan(1)=\frac{\pi}{4}
Inverse trig functions return angles in their restricted ranges: \arcsin in [-\pi/2,\pi/2]; \arccos in [0,\pi]; \arctan in (-\pi/2,\pi/2). These restrictions ensure the inverse is a function (one-to-one).

Example 2

hard
Simplify \sin(\arccos(x)) for x\in[-1,1] without trigonometric functions in the final answer.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
A right triangle has opposite side 5 and hypotenuse 13. Find the angle \theta opposite the side of length 5 using \arcsin.

Example 2

medium
Solve 2\sin(x)-1=0 on [0,2\pi] using \arcsin, and explain why there are two solutions.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

trigonometric functionsinverse functiondomain