Inverse circular functions: VCE Specialist Mathematics Unit 3

What this dot point is asking

VCAA wants you to define the inverse circular functions $\arcsin$, $\arccos$ and $\arctan$ (also written $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$), to state and justify the domain restrictions that make sine, cosine and tangent invertible, to know the exact domains and ranges of the inverses, to sketch their graphs, and to evaluate exact values and composite expressions. These functions then appear as antiderivatives and as the results of integration later in the course.

Why we restrict the domain

A function has an inverse only if it is one-to-one. Sine, cosine and tangent are periodic, so each horizontal line meets the full graph infinitely often. To create an invertible function we keep a single interval on which the function takes every output value exactly once.

• $\sin$ is restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, where it increases from $-1$ to $1$.
• $\cos$ is restricted to $[0, \pi]$, where it decreases from $1$ to $-1$.
• $\tan$ is restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, where it increases over all of $\mathbb{R}$.

These chosen intervals become the ranges of the inverse functions, which is why answers must lie inside them.

Domains, ranges and graphs

Each inverse graph is the reflection of the corresponding restricted function in the line $y = x$.

• $y = \arcsin x$ increases from $\left(-1, -\frac{\pi}{2}\right)$ through the origin to $\left(1, \frac{\pi}{2}\right)$.
• $y = \arccos x$ decreases from $(-1, \pi)$ through $\left(0, \frac{\pi}{2}\right)$ to $(1, 0)$.
• $y = \arctan x$ increases through the origin with horizontal asymptotes $y = \pm\frac{\pi}{2}$.

A useful identity is $\arcsin x + \arccos x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.

Examples in context

Example 1. $\arcsin(-1) = -\frac{\pi}{2}$ and $\arccos(-1) = \pi$, the endpoints of their ranges.

Example 2. Using the identity, $\arccos\frac{3}{5} = \frac{\pi}{2} - \arcsin\frac{3}{5}$.

Try this

Q1. State the domain and range of $y = \arctan x$. [2 marks]

• Cue. Domain $\mathbb{R}$, range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Q2. Find the exact value of $\arcsin\!\left(-\frac{\sqrt{2}}{2}\right)$. [2 marks]

• Cue. $-\frac{\pi}{4}$, since it lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Q3. Evaluate $\sin\!\left(\arccos\frac{5}{13}\right)$. [3 marks]

• Cue. With $\cos\theta = \frac{5}{13}$ and $\theta \in [0, \pi]$, $\sin\theta = \frac{12}{13}$.