Question

Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

Hint:

When evaluating inverse trigonometric functions, simplify the expression step by step using known values for sine and cosine.

Answer 1

We start by evaluating $ \cos^{-1} \frac{\sqrt{3}}{2}$. We know that $ \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}$. Now, substitute this value into the expression: \[ \tan^{-1} \left[ 2 \sin \left( 2 \times \frac{\pi}{6} \right) \right] = \tan^{-1} \left[ 2 \sin \left( \frac{\pi}{3} \right) \right] = \tan^{-1} \left[ 2 \times \frac{\sqrt{3}}{2} \right] = \tan^{-1} (\sqrt{3}) \] We know that \( \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \), so the final answer is: \[ \frac{\pi}{6} \]