3.9 Inverse Trigonometric FunctionsHL ONLY

1. Restricting Domains for Inverse Functions

Trigonometric functions fail the horizontal line test because they are periodic, meaning they lack inverse functions across their full domains. Establishing inverse functions requires restricting the domains to ensure a one-to-one mapping. These boundaries produce principal values.

The Inverse Sine ($\arcsin x$ or $\sin^{-1} x$):
The function $f(x) = \sin x$ is restricted to the domain $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
The inverse function $f^{-1}(x) = \arcsin x$ has:
  • Domain: $x \in [-1, 1]$
  • Range: $y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
x y 1 -1 π/2 -π/2 y = arcsin(x)
The Inverse Cosine ($\arccos x$ or $\cos^{-1} x$):
The function $f(x) = \cos x$ is restricted to the domain $[0, \pi]$.
The inverse function $f^{-1}(x) = \arccos x$ has:
  • Domain: $x \in [-1, 1]$
  • Range: $y \in [0, \pi]$
x y 1 -1 π π/2 y = arccos(x)
The Inverse Tangent ($\arctan x$ or $\tan^{-1} x$):
The function $f(x) = \tan x$ is restricted to the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
The inverse function $f^{-1}(x) = \arctan x$ has:
  • Domain: $x \in \mathbb{R}$
  • Range: $y \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
x y π/2 -π/2 y = arctan(x)

EXAMPLE 1 (Evaluating Exact Inverse Values)

We can use trigonometric values for exact evaluations. The expression $\arcsin(a)$ represents the principal solution to $\sin x = a$.

  • $\arcsin(0.5) = \dfrac{\pi}{6}$
  • $\arcsin(-0.5) = -\dfrac{\pi}{6}$
  • $\arccos(0.5) = \dfrac{\pi}{3}$
  • $\arccos(-0.5) = \dfrac{2\pi}{3}$
  • $\arctan(1) = \dfrac{\pi}{4}$
  • $\arctan(-1) = -\dfrac{\pi}{4}$
For boundary values: $\arcsin(0) = 0$, $\arccos(0) = \dfrac{\pi}{2}$, and $\arctan(0) = 0$.

2. Composition of Trigonometric Inverses

Composing a trigonometric function with its inverse yields the identity function, provided the values fall within the restricted domains:

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

Critical Exception:

The reverse relations, such as $\arcsin(\sin x) = x$, hold only if $x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

For example: $\arcsin\left(\sin \dfrac{5\pi}{6}\right) = \arcsin(0.5) = \dfrac{\pi}{6} \neq \dfrac{5\pi}{6}$.

EXAMPLE 2 (Applying Difference Identities)

Show that $\arctan 3 - \arctan 0.5 = \dfrac{\pi}{4}$.

Step 1: Assign variables to the inverse functions:
Let $A = \arctan 3$ and $B = \arctan 0.5$.
Step 2: Apply the tangent difference identity:
$$\begin{aligned} \tan(A - B) &= \dfrac{\tan A - \tan B}{1 + \tan A \tan B} \\ &= \dfrac{3 - 0.5}{1 + 3(0.5)} \\ &= \dfrac{2.5}{2.5} \\ &= 1 \end{aligned}$$
Step 3: The angle $(A - B)$ evaluates to either $\dfrac{\pi}{4}$ or $-\dfrac{3\pi}{4}$. Because $\arctan 3 > \arctan 0.5$, the difference must be positive.
Therefore, $A - B = \dfrac{\pi}{4} \implies \arctan 3 - \arctan 0.5 = \dfrac{\pi}{4}$.

EXAMPLE 3 (Right Triangle Models for Inverses)

Evaluate the expressions:
$A = \tan\left(\arctan\dfrac{2}{3}\right)$
$B = \sin\left(\arctan\dfrac{2}{3}\right)$
$C = \cos\left(\arctan\dfrac{2}{3}\right)$

The first evaluation uses inverse identity properties to isolate the argument:
$A = \mathbf{\dfrac{2}{3}}$.
For the remaining expressions, let $\theta = \arctan\dfrac{2}{3}$, yielding $\tan\theta = \dfrac{2}{3}$.
Construct a right triangle with an opposite side of $2$ and an adjacent side of $3$. By the Pythagorean theorem, the hypotenuse is:
$\text{Hypotenuse} = \sqrt{2^2 + 3^2} = \sqrt{13}$
Extracting the remaining ratios from this triangle yields:
  • $B = \sin\theta = \mathbf{\dfrac{2}{\sqrt{13}}}$
  • $C = \cos\theta = \mathbf{\dfrac{3}{\sqrt{13}}}$