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. These restricted boundaries produce what are known as 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 resulting inverse function $f^{-1}(x) = \arcsin x$ evaluates with:
  • 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 resulting inverse function $f^{-1}(x) = \arccos x$ evaluates with:
  • 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 strictly to the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
The resulting inverse function $f^{-1}(x) = \arctan x$ evaluates with:
  • 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 standard trigonometric values to calculate exact analytical evaluations. The expression $\arcsin(a)$ intrinsically represents the principal solution to the equation $\sin x = a$.

  • $\arcsin(0.5) = \dfrac{\pi}{6}$
  • $\arcsin(-0.5) = -\dfrac{\pi}{6}$
  • $\arccos(0.5) = \dfrac{\pi}{3}$
  • $\arccos(-0.5) = \dfrac{2\pi}{3}$
  • $\arctan(1) = \dfrac{\pi}{4}$
  • $\arctan(-1) = -\dfrac{\pi}{4}$
For absolute 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 direct inverse invariably yields the identity function, provided the specific values mathematically fall strictly within the restricted functional 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 natively formulated 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)

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

Step 1: Assign operative 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: Consequently, the angle sequence $(A - B)$ evaluates conceptually to either $\dfrac{\pi}{4}$ or $-\dfrac{3\pi}{4}$. Because conceptually $\arctan 3 > \arctan 0.5$, the differential must intrinsically 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 analytical 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 naturally utilizes direct inverse identity properties, isolating the argument purely:
$A = \mathbf{\dfrac{2}{3}}$.
For evaluating the remaining expressions analytically, establish the structural angle $\theta = \arctan\dfrac{2}{3}$, logically yielding $\tan\theta = \dfrac{2}{3}$.
Construct a standard right-angled geometric triangle mapping an opposite side magnitude of $2$ strictly corresponding to an adjacent base magnitude of $3$. Applying Pythagoras' theorem calculates the hypotenuse strictly as:
$\text{Hypotenuse} = \sqrt{2^2 + 3^2} = \sqrt{13}$
Extracting the remaining fundamental ratios reliably from this finalized geometric model firmly yields:
  • $B = \sin\theta = \mathbf{\dfrac{2}{\sqrt{13}}}$
  • $C = \cos\theta = \mathbf{\dfrac{3}{\sqrt{13}}}$