Inverse trigonometric (circular) functions (WACE Year 12 Mathematics Specialist Unit 3)

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.

Compositions with a right triangle

A common task is to simplify an expression such as $\cos(\sin^{-1} x)$ into an algebraic form. Set $\theta = \sin^{-1} x$, so $\sin\theta = x$ with $\theta$ in the first or fourth quadrant. Draw a right triangle with opposite side $x$ and hypotenuse $1$; the adjacent side is $\sqrt{1 - x^2}$. Then $\cos(\sin^{-1} x) = \sqrt{1 - x^2}$, taking the non-negative root because $\cos\theta \ge 0$ on the range of $\sin^{-1}$. The same triangle gives $\tan(\sin^{-1} x) = \dfrac{x}{\sqrt{1 - x^2}}$. This technique turns nested trig-and-inverse-trig expressions into clean surds and is frequently tested.

Derivatives that motivate the inverses

Although the differentiation itself belongs to the calculus dot point, the inverse circular functions matter precisely because they have clean derivatives: $\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$, $\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1 - x^2}}$ and $\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}$. Read in reverse, these are antiderivatives that appear constantly in Unit 4 integration, so the domains and ranges here underpin a great deal of later work.

Symmetry relationships

The three inverses are linked by useful identities. Because sine and cosine are co-functions, $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$ for every $x \in [-1, 1]$, which often shortens an evaluation. Both $\sin^{-1}$ and $\tan^{-1}$ are odd functions, so $\sin^{-1}(-x) = -\sin^{-1} x$ and $\tan^{-1}(-x) = -\tan^{-1} x$, while $\cos^{-1}$ satisfies $\cos^{-1}(-x) = \pi - \cos^{-1} x$ instead, reflecting its range $[0, \pi]$.