Inverse Trigonometric Functions

In short: The inverse trig functions find an angle from a ratio, with a restricted range so each gives one answer. $sin^{- 1} x$ and $cos^{- 1} x$ take $x$ from $- 1$ to $1$ , with ranges $[- \frac{π}{2}, \frac{π}{2}]$ and $[0, π]$ , and $tan^{- 1} x$ takes any $x$ .

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

Year 12Using a triangle

Find $cos (sin^{- 1} \frac{3}{5})$ exactly.

Let $θ = sin^{- 1} \frac{3}{5}$ , so $sin θ = \frac{3}{5}$ with $θ$ acute. Draw a right triangle with opposite 3 and hypotenuse 5. The adjacent side is 4, so the answer is $\frac{4}{5}$ .

Double angles. Let $θ = tan^{- 1} \frac{1}{3}$ , so $tan θ = \frac{1}{3}$ . Then $cos 2 θ = \frac{1 - tan ^{2} θ}{1 + tan ^{2} θ} = \frac{8/9}{10/9} = \frac{4}{5}$ . And $sin (2 sin^{- 1} x) = 2 sin θ cos θ = 2 x 1 - x^{2}$ with $θ = sin^{- 1} x$ (the root is positive because $θ$ is between $- \frac{π}{2}$ and $\frac{π}{2}$ ).

Sums of inverse tangents. For $tan^{- 1} 1 + tan^{- 1} 2$ : the tan of the sum is $\frac{1 + 2}{1 - 2} = - 3$ . Each angle is between $0$ and $\frac{π}{2}$ , and $tan^{- 1} 2 > \frac{π}{4}$ , so the sum is between $\frac{π}{2}$ and $π$ . The angle there with tangent $- 3$ is $π - tan^{- 1} 3$ .

Equations. Use $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$ . If $sin^{- 1} x = cos^{- 1} x$ , each is $\frac{π}{4}$ , so $x = \frac{1}{2}$ . Similarly $tan^{- 1} x + tan^{- 1} \frac{1}{x} = \frac{π}{2}$ for $x > 0$ and $- \frac{π}{2}$ for $x < 0$ .

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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Try 3 questions now

Find $sin^{- 1} \frac{3}{2}$ .
1. A $\frac{π}{6}$
2. B $\frac{π}{3}$
3. C $\frac{2 π}{3}$
4. D $\frac{π}{4}$
Show answer
B: $\frac{π}{3}$
$sin \frac{π}{3} = \frac{3}{2}$ and it is in the range.
Find $cos^{- 1} 0$ .
1. A 0
2. B $\frac{π}{2}$
3. C $π$
4. D $- \frac{π}{2}$
Show answer
B: $\frac{π}{2}$
$cos \frac{π}{2} = 0$ .
Find $tan^{- 1} 1$ .
1. A $\frac{π}{4}$
2. B $\frac{π}{2}$
3. C 1
4. D $\frac{3 π}{4}$
Show answer
A: $\frac{π}{4}$
$tan \frac{π}{4} = 1$ .