Q.State True or False: All trigonometric functions have inverse over their respective domains.

Concept: Inverse Function Relationship — A function has an inverse only if it is bijective (one-to-one and onto). Trigonometric functions are periodic, so they are not one-to-one over their entire natural domain.

Step 1: For a function to have an inverse, each output must correspond to exactly one input. Trigonometric functions like $\sin x$, $\cos x$, and $\tan x$ repeat values over their full domains (e.g., $\sin 0 = \sin \pi = 0$), so they are not one-to-one.

Step 2: To define inverses, we restrict each trigonometric function to a specific interval where it becomes bijective — for example, $\sin x$ on $[-\pi/2, \pi/2]$, $\cos x$ on $[0, \pi]$, $\tan x$ on $(-\pi/2, \pi/2)$. These restricted domains give us the principal value branches. …

The key idea here is the definition of an inverse function. A function $f$ has an inverse $f^{-1}$ if and only if $f$ is bijective — that is, both one-to-one (injective) and onto (surjective). For a function to be one-to-one, every horizontal line must intersect its graph at most once (the Horizontal Line Test).

Now look at any trigonometric function, say $\sin x$. Its graph is a wave that repeats every $2\pi$. Over its natural domain $\mathbb{R}$, the horizontal line $y = 0.5$ hits the sine curve infinitely many times. So $\sin x$ is not one-to-one on $\mathbb{R}$, and therefore it cannot have an inverse over $\mathbb{R}$.

The same is true for $\cos x$, $\tan x$, and all the other trig functions. Each one is periodic, so each fails the Horizontal Line Test over its full domain.

1. Why the confusion arises: In practice, we do talk about $\sin^{-1} x$, $\cos^{-1} x$, etc. But these are defined only after we restrict the domain of the original function to an interval where it is one-to-one. For $\sin x$, the standard restriction is $[-\pi/2, \pi/2]$; for $\cos x$, it's $[0, \pi]$; for $\tan x$, it's $(-\pi/2, \pi/2)$. These restricted functions are one-to-one and onto their ranges, so they have inverses.

2. The critical distinction: The question says "over their respective domains." The phrase "respective domains" here means the natural domains of the functions — $\mathbb{R}$ for sine and cosine, $\mathbb{R}$ minus odd multiples of $\pi/2$ for tangent, etc. Over these full domains, no trigonometric function is one-to-one. …

Method: Deciding whether a function is invertible on a given domain

For any "does the inverse exist?" claim, test one-to-one-ness, not the notation.

Steps

Step 1: Apply the one-to-one test.

A function is invertible only if it is one-to-one (passes the horizontal line test) on the stated domain.

Step 2: Check periodicity. …

Common Mistakes

Mistake 1: Believing $\sin^{-1}$, $\cos^{-1}$, etc. invert the trig functions on their whole domains.

Why it's wrong: those symbols invert the RESTRICTED functions, not $\sin x$ or $\cos x$ over all of $\mathbb{R}$. Correct approach: an inverse over the full periodic domain does not exist, so the statement is False. …