Q.Find the principal value of the following: $\tan^{-1} \left[2\cos \left(2\sin^{-1} \frac{1}{2}\right)\right]$

Concept: Inverse Trigonometric Graphs — we evaluate from the inside out, using known standard angles.

1. First, $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ (principal value range for $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$).
2. Then $2\sin^{-1} \frac{1}{2} = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}$.
3. So $\cos\left(2\sin^{-1} \frac{1}{2}\right) = \cos\frac{\pi}{3} = \frac{1}{2}$. …

Concept and Intuition

When you see a nested expression like $\tan^{-1}[2\cos(2\sin^{-1}\frac12)]$, the natural instinct is to work from the inside out. The key is to remember that inverse trigonometric functions return angles (principal values), not ratios. So $\sin^{-1}\frac12$ is an angle whose sine is $\frac12$ — and we know exactly which angle that is in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Once you have that angle, the rest is just ordinary trigonometry: double the angle, take its cosine, multiply by 2, and then ask: what angle has this number as its tangent? The final step must respect the principal value branch of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Let's walk through it.

1. Evaluate the innermost inverse sine

   $\sin^{-1}\frac12$ asks: which angle $\theta$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ has $\sin\theta = \frac12$?

   The standard angle is $\frac{\pi}{6}$ (30°). So:

2. Substitute into the cosine expression

   The argument of the cosine becomes $2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

   So we need:

$\cos\frac{\pi}{3} = \frac12$, therefore:

3. Now evaluate the outer inverse tangent

   The problem reduces to:

   $\tan^{-1}(1)$ …

Method: Evaluating nested inverse and direct trig functions

Use this for expressions where an inverse function sits inside a direct one (or vice versa); always work from the innermost bracket outward.

Steps

Step 1: Evaluate the innermost inverse function

Compute $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ first, using its principal branch.

Step 2: Substitute upward one layer at a time …

Common Mistakes

Mistake 1: Working from the outside in

Why it's wrong: the outer $\tan^{-1}$ cannot be applied until its argument is a number; starting outside leaves undefined expressions. Correct approach: evaluate $\sin^{-1}\frac{1}{2}$ first, then move outward.

Mistake 2: Dropping the coefficients (the two 2's) …