Q.$\sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to (A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $1$

Concept: Inverse Trigonometric Graphs — the range of $\sin^{-1}x$ is $[-\pi/2, \pi/2]$, so we must pick the principal value.

Step 1: Evaluate the inner inverse.

$\sin^{-1}(-1/2) = -\pi/6$, because $\sin(-\pi/6) = -1/2$ and $-\pi/6$ lies in the principal range.

Step 2: Substitute into the expression. …

We need to evaluate $\sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)\right)$. The trick here is not to rush into computation — first, understand what $\sin^{-1}(-1/2)$ means.

Inverse sine graph intuition: The principal value branch of $\sin^{-1}x$ lies in $[-\pi/2, \pi/2]$. For a negative input, the output is a negative angle in that interval. So $\sin^{-1}(-1/2)$ is not $-\pi/6$? Let’s check: $\sin(-\pi/6) = -1/2$, and $-\pi/6$ is indeed in $[-\pi/2, \pi/2]$. So yes, $\sin^{-1}(-1/2) = -\pi/6$.

But there’s a more general principle at work here — one that saves time and avoids sign errors.

Now substitute:

1. Rewrite the expression

2. Add the angles …

Method: Evaluating a trig function applied to an inverse-trig expression

Use this whenever a plain trig function wraps an angle built from an inverse-trig term, e.g. $\sin(c - \sin^{-1} a)$.

Steps

Step 1: Evaluate the inner inverse term as a concrete angle.

Compute the inverse-trig value first, respecting its principal range. For a negative argument, remember $\sin^{-1}$ is odd: $\sin^{-1}(-a) = -\sin^{-1}(a)$, so the result is a negative angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 2: Substitute and combine the angles. …

Common Mistakes

Mistake 1: Evaluating $\sin^{-1}\left(-\frac{1}{2}\right)$ as $-\frac{\pi}{3}$ or $\frac{7\pi}{6}$.

Why it's wrong: the correct principal value is $-\frac{\pi}{6}$, since $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$ and $-\frac{\pi}{6}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Correct approach: use $\sin^{-1}(-x) = -\sin^{-1}x$ to get $-\frac{\pi}{6}$.

Mistake 2: Mishandling the double negative. …