Inverse Trigonometric Functions: Edge Cases and Subtle Traps (3)

Question

Evaluate $\sin^{-1}\left(\sin\dfrac{5\pi}{6}\right)$.

Solution — Step by Step

Why This Works

The function $\sin^{-1}$ is the inverse only when restricted to the principal range $[-\pi/2, \pi/2]$, where $\sin$ is one-to-one. For angles outside this range, $\sin^{-1}(\sin\theta) \neq \theta$ in general. We must convert the angle to an equivalent one inside the principal range using identities like $\sin(\pi - x) = \sin x$ or $\sin(2\pi + x) = \sin x$.

This is a JEE Main and JEE Advanced favourite trap because the formula $\sin^{-1}(\sin x) = x$ is true only conditionally. Students who memorise without thinking get it wrong.

Alternative Method

Picture the unit circle. The point at angle $5\pi/6$ has the same $y$-coordinate as the point at angle $\pi/6$ (both are at $y = 1/2$). Since $\sin^{-1}$ asks “which angle in $[-\pi/2, \pi/2]$ has this $y$?”, the answer is $\pi/6$, not $5\pi/6$.

Common Mistake

For $\cos^{-1}(\cos x)$, the principal range is $[0, \pi]$, not $[-\pi/2, \pi/2]$. So $\cos^{-1}(\cos(7\pi/6)) = \cos^{-1}(\cos(7\pi/6 - 2\pi)) = \cos^{-1}(\cos(-5\pi/6)) = \cos^{-1}(\cos(5\pi/6)) = 5\pi/6$. Each inverse trig has its own principal range — memorise them separately.