3.9 Inverse Trigonometric Functions (HL)

1. Restricting Domains for Inverse Functions

Standard trigonometric functions inherently fail the horizontal line test due to their periodic nature, meaning they do not possess inverse functions across their full domains. Establishing valid inverse functions requires rigorously restricting the initial domains to ensure a strictly one-to-one mapping.

The Inverse Sine ($\arcsin x$ or $\sin^{-1} x$):
The function $f(x) = \sin x$ is restricted to the domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
The resulting inverse function $f^{-1}(x) = \arcsin x$ evaluates with Domain: $x \in [-1, 1]$ and Range: $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
The Inverse Cosine ($\arccos x$ or $\cos^{-1} x$):
The function $f(x) = \cos x$ is restricted to the domain $[0, \pi]$.
The resulting inverse function $f^{-1}(x) = \arccos x$ evaluates with Domain: $x \in [-1, 1]$ and Range: $y \in [0, \pi]$.
The Inverse Tangent ($\arctan x$ or $\tan^{-1} x$):
The function $f(x) = \tan x$ is restricted strictly to the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
The resulting inverse function $f^{-1}(x) = \arctan x$ evaluates with Domain: $x \in \mathbb{R}$ and Range: $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

EXAMPLE 1 (Evaluating Inverse Values)

Utilizing standard trigonometric values allows for exact inverse evaluations. For instance, $\arcsin(a)$ represents the principal solution to the equation $\sin x = a$.

  • $\arcsin(0.5) = \frac{\pi}{6}$
  • $\arcsin(-0.5) = -\frac{\pi}{6}$
  • $\arccos(0.5) = \frac{\pi}{3}$
  • $\arccos(-0.5) = \frac{2\pi}{3}$
  • $\arctan(1) = \frac{\pi}{4}$
  • $\arctan(-1) = -\frac{\pi}{4}$

For boundary values: $\arcsin(0) = 0$, $\arccos(0) = \frac{\pi}{2}$, and $\arctan(0) = 0$.

2. Composition of Trigonometric Inverses

Composing a trigonometric function with its direct inverse yields the identity function, provided the input falls strictly within the restricted functional domain:

  • $\sin(\arcsin x) = x$ (valid for $x \in [-1, 1]$)
  • $\cos(\arccos x) = x$ (valid for $x \in [-1, 1]$)
  • $\tan(\arctan x) = x$ (valid for $x \in \mathbb{R}$)

Critical Exception: The reverse composition $\arcsin(\sin x) = x$ holds only if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. For example, $\arcsin(\sin\frac{5\pi}{6}) = \arcsin(0.5) = \frac{\pi}{6} \ne \frac{5\pi}{6}$.

EXAMPLE 2

Demonstrate algebraically that $\arctan 3 - \arctan 0.5 = \frac{\pi}{4}$.

Assign variables: $A = \arctan 3$ and $B = \arctan 0.5$.
Apply the tangent difference identity: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Substitute known values: $\tan(A - B) = \frac{3 - 0.5}{1 + 3(0.5)} = \frac{2.5}{2.5} = 1$.
Consequently, $A - B$ evaluates to either $\frac{\pi}{4}$ or $-\frac{3\pi}{4}$. Because $\arctan 3 > \arctan 0.5$, the difference must be positive, confirming $A - B = \mathbf{\frac{\pi}{4}}$.

EXAMPLE 3

Evaluate the expressions $A = \tan(\arctan\frac{2}{3})$, $B = \sin(\arctan\frac{2}{3})$, and $C = \cos(\arctan\frac{2}{3})$.

The first evaluation utilizes direct identity properties: $A = \mathbf{\frac{2}{3}}$.
For the remaining expressions, establish $\theta = \arctan\frac{2}{3}$, yielding $\tan\theta = \frac{2}{3}$.
Construct a right-angled triangle with an opposite side of $2$ and an adjacent side of $3$. Applying Pythagoras' theorem calculates the hypotenuse as $\sqrt{2^2 + 3^2} = \sqrt{13}$.
Extracting the ratios from this geometric model yields:
$B = \sin\theta = \mathbf{\frac{2}{\sqrt{13}}}$
$C = \cos\theta = \mathbf{\frac{3}{\sqrt{13}}}$.