Understanding Inverse Trigonometric Functions in Maths

Inverse trigonometric functions are multivalued inverses of the basic trigonometric functions, defined with principal branches to ensure they are functions in the strict mathematical sense.

Mathematical Formulation of Inverse Trigonometric Functions and Their Principal Branches

Let $y = \sin^{-1} x$ denote the inverse sine function. The relation $\sin y = x$ holds, but since $\sin y$ is periodic, the equation has infinitely many solutions for $y \in \mathbb{R}$ for each $x \in [-1,1]$. To define $\sin^{-1} x$ as a function, the range is restricted to $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Let $y = \cos^{-1} x$ denote the inverse cosine function. The relation $\cos y = x$ is satisfied, and by restricting $y$ to $[0, \pi]$, $\cos^{-1} x$ becomes a well-defined function for $x \in [-1, 1]$.

For $y = \tan^{-1} x$, the inverse tangent function is defined such that $\tan y = x$ and $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, ensuring uniqueness for every $x \in \mathbb{R}$.

Similarly, for $y = \cot^{-1} x$, by considering $\cot y = x$ and restricting $y \in (0, \pi)$, $\cot^{-1} x$ is defined for $x \in \mathbb{R}$.

Let $y = \sec^{-1} x$ denote the inverse secant function. The domain is $|x| \geq 1$ and the principal value branch is chosen as $y \in [0, \pi] \setminus \{\frac{\pi}{2}\}$.

Let $y = \csc^{-1} x$ denote the inverse cosecant function. The domain is $|x| \geq 1$, and the principal value is chosen from $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$.

Domain and Principal Value Ranges of Inverse Trigonometric Functions

For $y = \sin^{-1} x$: The domain is $x \in [-1,1]$; the principal value lies in $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

For $y = \cos^{-1} x$, the domain is $x \in [-1,1]$, with range $y \in [0, \pi]$.

For $y = \tan^{-1} x$, the domain is $x \in \mathbb{R}$ and the range is $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

For $y = \cot^{-1} x$, the domain is $x \in \mathbb{R}$, principal value $y \in (0,\pi)$.

For $y = \sec^{-1} x$, the domain is $|x| \geq 1$ and the range is $y \in [0,\pi] \setminus \{\frac{\pi}{2}\}$.

For $y = \csc^{-1} x$, the domain is $|x| \geq 1$, range $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$.

Functional Properties and Symmetries of Inverse Trigonometric Functions

The following symmetry properties hold:

$\sin^{-1}(-x) = -\sin^{-1} x$ for $x \in [-1,1]$.

$\cos^{-1}(-x) = \pi - \cos^{-1} x$ for $x \in [-1,1]$.

$\tan^{-1}(-x) = -\tan^{-1} x$ for $x \in \mathbb{R}$.

$\cot^{-1}(-x) = \pi - \cot^{-1} x$ for $x \in \mathbb{R}$.

$\sec^{-1}(-x) = \pi - \sec^{-1} x$ for $|x| \geq 1$.

$\csc^{-1}(-x) = -\csc^{-1} x$ for $|x| \geq 1$.

Inter-Relations among Inverse Trigonometric Functions

For $x \in [-1,1] \setminus \{0\}$, $\sin^{-1} x = \csc^{-1} \left(\dfrac{1}{x}\right)$.

For $x \in [-1,1] \setminus \{0\}$, $\cos^{-1} x = \sec^{-1} \left( \dfrac{1}{x} \right)$.

For $x \in \mathbb{R} \setminus \{0\}$, $\tan^{-1} x = \cot^{-1} \left( \dfrac{1}{x} \right)$ if $x>0$, and $\tan^{-1} x = \cot^{-1} \left( \dfrac{1}{x} \right) - \pi$ if $x<0$.

For $x \in \mathbb{R} \setminus \{0\}$, $\cot^{-1} x = \tan^{-1} \left( \dfrac{1}{x} \right)$ if $x>0$, and $\cot^{-1} x = \tan^{-1} \left( \dfrac{1}{x} \right) + \pi$ if $x<0$.

Fundamental Identities Involving Inverse Trigonometric Functions

For $x \in [-1,1]$, $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$. This may be verified as follows.

Let $y = \sin^{-1} x \implies x = \sin y$, where $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Then, $\cos^{-1} x = \cos^{-1} (\sin y)$. Let $z = \cos^{-1} (\sin y)$, then $\cos z = \sin y$. But $\cos z = \sin \left( \dfrac{\pi}{2} - z \right )$, so $\sin y = \sin \left( \dfrac{\pi}{2} - z \right )$.

Therefore, within the principal value intervals, $y = \dfrac{\pi}{2} - z \implies z = \dfrac{\pi}{2} - y$.

Thus, $\sin^{-1} x + \cos^{-1} x = y + z = y + \dfrac{\pi}{2} - y = \dfrac{\pi}{2}$.

Similarly, for $x \in \mathbb{R}$, $\tan^{-1} x + \cot^{-1} x = \dfrac{\pi}{2}$.

Reduction to Algebraic Expressions — Double and Triple Angle Formulas

For $x \in [-1,1]$, $2\sin^{-1} x = \sin^{-1} (2x\sqrt{1 - x^2})$, with derivation as follows.

Let $\theta = \sin^{-1} x$, with $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

Then $2\theta = 2\sin^{-1} x$. From the double angle formula for sine, $\sin 2\theta = 2\sin \theta \cos \theta$.

$\sin \theta = x$, $\cos \theta = \sqrt{1 - x^2}$, since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ ensures $\cos \theta \geq 0$.

Thus, $\sin 2\theta = 2x\sqrt{1 - x^2}$, so $2\sin^{-1} x = \sin^{-1}(2x\sqrt{1 - x^2})$ for $x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$.

Triple angle and other similar reductions follow analogously, with explicit evaluation via substitution of the principal values.

Algebraic Relationships Among Compositions of Inverse Trigonometric Functions

If $-1 \leq x \leq 1$, then $\sin(\sin^{-1} x) = x$. If $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$, then $\sin^{-1}(\sin x) = x$.

If $-1 \leq x \leq 1$, then $\cos(\cos^{-1} x) = x$. If $0 \leq x \leq \pi$, then $\cos^{-1}(\cos x) = x$.

If $x \in \mathbb{R}$, then $\tan(\tan^{-1} x) = x$. If $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$, then $\tan^{-1}(\tan x) = x$.

Principal Value Evaluation: Solved Examples

Example: Evaluate $\cos^{-1} \left(\dfrac{1}{2}\right) + 2\sin^{-1} \left( \dfrac{1}{2} \right)$.

Let $y_1 = \cos^{-1} \left( \dfrac{1}{2} \right)$.

$\cos y_1 = \dfrac{1}{2}$

$y_1 = \dfrac{\pi}{3}$, since $0 \leq y_1 \leq \pi$.

Let $y_2 = \sin^{-1} \left( \dfrac{1}{2} \right)$.

$\sin y_2 = \dfrac{1}{2}$

$y_2 = \dfrac{\pi}{6}$, since $-\dfrac{\pi}{2} \leq y_2 \leq \dfrac{\pi}{2}$.

Thus, the sum is $y_1 + 2y_2 = \dfrac{\pi}{3} + 2 \cdot \dfrac{\pi}{6} = \dfrac{\pi}{3} + \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.

For further analysis of inverse trigonometric relations and their role in calculus, see Integral Calculus Concepts.

Common Exam Errors in Inverse Trigonometric Functions

A frequent error is confusing $\sin^{-1} x$ with $(\sin x)^{-1}$; the former denotes the inverse function, while the latter represents the reciprocal.

Omitting the principal branch restrictions leads to incorrect principal values, especially in compositions such as $\sin(\sin^{-1} x)$.

Misidentification of domains for expressions like $\sec^{-1} x$ or $\csc^{-1} x$ is a critical error. Ensure $|x| \geq 1$ for these functions.

For a structured overview of these functions, refer to Inverse Trigonometric Functions In Maths.