Q.State True or False: The principal value of $\sin^{-1}\left(\cos\left(\sin^{-1}\frac{1}{2}\right)\right)$ is $\frac{\pi}{3}$.

Concept: Inverse Sine Principal Value – The principal value of $\sin^{-1}x$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Step 1: Evaluate the innermost term.

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

Step 2: Substitute and simplify.

$\cos\left(\sin^{-1}\frac{1}{2}\right) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$.

Step 3: Now find $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$. …

Concept and Intuition

When you see a nested inverse trigonometric function like $\sin^{-1}(\cos(\sin^{-1}\frac{1}{2}))$, the natural instinct is to work from the innermost layer outward. But there's a subtle trap: inverse trigonometric functions have restricted principal value ranges. For $\sin^{-1}x$, the output lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For $\cos^{-1}x$, it lies in $[0, \pi]$. Here, we only have $\sin^{-1}$, so we must ensure every angle we produce falls within its principal range.

The problem asks whether the entire expression simplifies to $\frac{\pi}{3}$. Let's verify step by step.

Step-by-Step Solution

1. Evaluate the innermost term: $\sin^{-1}\frac{1}{2}$

We know $\sin\frac{\pi}{6} = \frac{1}{2}$, and $\frac{\pi}{6}$ lies in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ of $\sin^{-1}$. Therefore:

2. Now evaluate $\cos\left(\frac{\pi}{6}\right)$

We have:

So the expression becomes:

3. Evaluate $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$ …

Method: Evaluate nested inverse-trig expressions inside-out

Steps

Step 1: Evaluate the innermost inverse function using its principal range: $\sin^{-1}\tfrac12=\tfrac{\pi}{6}$.

Step 2: Apply the next (ordinary) function: $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$. …

Common Mistakes

Mistake 1: Using a non-principal value such as $\sin^{-1}\frac12=\frac{5\pi}{6}$.

Why it's wrong: $\sin^{-1}$ must return an angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so only $\frac{\pi}{6}$ qualifies. Correct approach: take the principal value at each layer.

Mistake 2: Confusing the middle $\cos\frac{\pi}{6}$ with an inverse cosine. …