1.05h–iPureStretch

Inverse Trigonometric Functions

Sin, cos and tan are many-to-one, so to invert them we restrict their domains. The resulting inverse functions arcsin, arccos and arctan each have a specific domain and range you must know.

30 min Trigonometry

The big picture

The principal ranges are $[-\pi/2,\pi/2]$ for arcsin, $[0,\pi]$ for arccos, and $(-\pi/2,\pi/2)$ for arctan. Sine and cosine inverses have input domain $[-1,1]$ ; arctan accepts every real input. An inverse trig key returns one principal value. Solving an equation on a wider interval still requires symmetry and periodicity. The graph of each inverse is the reflection of its restricted original graph in y = x.

Edexcel A level Maths: 6.5 Inverse Trigonometric FunctionsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Understand why the domains are restricted
State the domain and range of each inverse
Sketch arcsin, arccos, arctan
Evaluate inverse trig values

Restricting the domain

A function only has an inverse if it is one-to-one. We restrict $\sin$ to $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ , $\cos$ to $[0, \pi]$ , and $\tan$ to $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ . The inverse swaps domain and range.

\arcsin: [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]

Domain and range of arcsin.

Domains and ranges

$\arcsin x$ : domain $[-1,1]$ , range $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ . $\arccos x$ : domain $[-1,1]$ , range $[0,\pi]$ . $\arctan x$ : domain all reals, range $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ .

Example 1Find

\arcsin\tfrac12

1We need the angle in

[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]

with sine

\tfrac12

2That is

\tfrac{\pi}{6}

Answer

\tfrac{\pi}{6}

Tip — arccos has range [0, π] — its answers are never negative.

Principal values and the full solution set

The principal ranges are $[-\pi/2,\pi/2]$ for arcsin, $[0,\pi]$ for arccos, and $(-\pi/2,\pi/2)$ for arctan. Sine and cosine inverses have input domain $[-1,1]$ ; arctan accepts every real input.

An inverse trig key returns one principal value. Solving an equation on a wider interval still requires symmetry and periodicity. The graph of each inverse is the reflection of its restricted original graph in y = x.

Example 1Evaluate

\arcsin(\sin(5\pi/6))

, then solve

\sin x=1/2

[0,2\pi)

\sin(5\pi/6)=1/2

2The principal arcsin value of

1/2

\pi/6

, not

5\pi/6

3The full equation has two solutions,

\pi/6

and

5\pi/6

, in the stated interval.

AnswerPrincipal value

\pi/6

; equation solutions

\pi/6,5\pi/6

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Giving arccos a negative answer.

arccos has range [0, π]; answers are in that interval.

Using domain [−1,1] for arctan.

arctan accepts all real numbers.

Key results to remember

\arcsin x: [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]

Inverse sine.

\arccos x: [-1,1] \to [0, \pi]

Inverse cosine.

\arctan x: \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2})

Inverse tangent.

Key takeaways

Restrict domains to make sin, cos, tan one-to-one.
arcsin/arccos domain [−1,1]; arctan domain all reals.
Ranges: arcsin [−π/2, π/2], arccos [0, π], arctan (−π/2, π/2).

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

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