11. Inverse Trigonometric Functions | Mathematics Extension 1

Starting point: what you already know

Inverse functions (Year 11): $f^{- 1}$ undoes $f$ , exists only for one-to-one functions, and its graph is the reflection of $f$ in $y = x$ . Domain and range swap.
Exact values on the unit circle, for example $sin \frac{π}{6} = \frac{1}{2}$ , $cos \frac{π}{4} = \frac{1}{2}$ , $tan \frac{π}{3} = 3$ .
The graphs of $sin x$ , $cos x$ and $tan x$ , in radians.

The problem

$sin x = \frac{1}{2}$ has infinitely many solutions: $\frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}, \dots$ So $sin x$ is not one-to-one, and it has no inverse on all real numbers.

The fix is to restrict the domain to one piece where the function goes through every possible value exactly once.

Function	Restricted domain	So the inverse has domain	and range
$sin x$	$- \frac{π}{2} \leq x \leq \frac{π}{2}$	$- 1 \leq x \leq 1$	$- \frac{π}{2} \leq y \leq \frac{π}{2}$
$cos x$	$0 \leq x \leq π$	$- 1 \leq x \leq 1$	$0 \leq y \leq π$
$tan x$	$- \frac{π}{2} < x < \frac{π}{2}$	all real $x$	$- \frac{π}{2} < y < \frac{π}{2}$

What the symbols mean

$sin^{- 1} x$ is the angle between $- \frac{π}{2}$ and $\frac{π}{2}$ whose sine is $x$ . It is also written $arcsin x$ .

$cos^{- 1} x$ is the angle between 0 and $π$ whose cosine is $x$ . $tan^{- 1} x$ is the angle strictly between $- \frac{π}{2}$ and $\frac{π}{2}$ whose tangent is $x$ .

The graphs

Reflect each restricted piece in $y = x$ .

%%FIG:arcsincos%%

%%FIG:arctan%%

Key properties

Do and undo. For $- 1 \leq x \leq 1$ , $sin (sin^{- 1} x) = x$ always. But $sin^{- 1} (sin x) = x$ only when $x$ is in $[- \frac{π}{2}, \frac{π}{2}]$ .

Inverse trig functions

Why restrict

sin, cos, tan not one-to-one

sin⁻¹ x

Domain -1 to 1
Range -π/2 to π/2

cos⁻¹ x

Domain -1 to 1
Range 0 to π

tan⁻¹ x

Any x
Range between -π/2 and π/2

Properties

sin⁻¹(-x) = -sin⁻¹ x
cos⁻¹(-x) = π - cos⁻¹ x
sin⁻¹ x + cos⁻¹ x = π/2

Quick summary

$sin^{- 1} x$ : domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ .
$cos^{- 1} x$ : domain $[- 1, 1]$ , range $[0, π]$ .
$tan^{- 1} x$ : domain all real $x$ , range strictly between $- \frac{π}{2}$ and $\frac{π}{2}$ .
$sin^{- 1} (- x) = - sin^{- 1} x$ , $cos^{- 1} (- x) = π - cos^{- 1} x$ , $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$ .
$sin^{- 1} (sin x) = x$ only inside the range of $sin^{- 1}$ .

Quiz

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