What is wrong sign branch for arctan?

Tangent is odd, so tan^-1(-x) = -tan^-1(x); keep the result in (-π2, π2).

WASpecialist MathematicsSyllabus dot point

How do we restrict the circular functions so that their inverses are genuine functions, and what do those inverses look like?

Define the inverse circular functions with their restricted domains and ranges and sketch their graphs

WACE Specialist Unit 3 inverse circular functions: why sine, cosine and tangent must be domain-restricted to be invertible, the principal domains and ranges of arcsin, arccos and arctan, their graphs as reflections, and exact values, with a worked example.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Why a restriction is needed
The graphs as reflections
Evaluating exact values

What this dot point is asking

SCSA wants you to know why the restrictions are needed, state the domain and range of each inverse, sketch them, and evaluate exact values.

Why a restriction is needed

A function has an inverse only if it is one-to-one. The circular functions are periodic, so each output is hit infinitely often; no inverse exists on the full domain. We therefore restrict each to a largest interval on which it is one-to-one and covers the full range.

The graphs as reflections

Because the inverse of a function is its reflection in $y = x$ , each inverse graph comes from reflecting the appropriately restricted circular function. So $\sin^{-1}$ rises from $(-1, -\tfrac{\pi}{2})$ to $(1, \tfrac{\pi}{2})$ ; $\cos^{-1}$ falls from $(-1, \pi)$ to $(1, 0)$ ; and $\tan^{-1}$ increases through the origin with horizontal asymptotes at $y = \pm\tfrac{\pi}{2}$ .

Evaluating exact values

To evaluate, ask which angle in the principal range has the required circular value. For example $\cos^{-1}\!\left(-\tfrac{1}{2}\right)$ asks for the angle in $[0, \pi]$ with cosine $-\tfrac{1}{2}$ , which is $\tfrac{2\pi}{3}$ . The answer must always lie in the stated range.

Worked example

Evaluate $\sin^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)$ and $\tan^{-1}(-1)$

For $\sin^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)$ , find the angle in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ whose sine is $\tfrac{\sqrt{3}}{2}$ . That angle is $\tfrac{\pi}{3}$ .

For $\tan^{-1}(-1)$ , find the angle in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ whose tangent is $-1$ . That is $-\tfrac{\pi}{4}$ (the negative branch, since tangent is odd).

So $\sin^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right) = \tfrac{\pi}{3}$ and $\tan^{-1}(-1) = -\tfrac{\pi}{4}$ .