Evaluate: tan-1{2sin(4cos-1(√3/2))}
Solution:
Given:
\[ y = \tan^{-1}\left(2\sin\left(4\cos^{-1}\frac{\sqrt{3}}{2}\right)\right) \]
Step 1: Evaluate cos-1(√3/2)
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]
Step 2: Substitute
\[ y = \tan^{-1}\left(2\sin\left(4 \cdot \frac{\pi}{6}\right)\right) = \tan^{-1}\left(2\sin\left(\frac{2\pi}{3}\right)\right) \]
Step 3: Evaluate sin(2π/3)
\[ \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2} \]
So,
\[ 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \]
Step 4: Final evaluation
\[ y = \tan^{-1}(\sqrt{3}) \]
Since principal value range of tan-1(x) is \((- \pi/2, \pi/2)\),
\[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Final Answer:
Principal Value = \[ \frac{\pi}{3} \]