Question

Write the value of $ta{n}^{-1}\left[2\sin \right(2{\cos }^{-1}\frac{\sqrt{3}}{2}\left)\right]$.

Answer 1

First solve the inside bracket value $2co{s}^{-1}\frac{\sqrt{3}}{2}$

The range of the principal value of $co{s}^{-1}x$ is $[0,\pi ]$

Let $co{s}^{-1}\frac{\sqrt{3}}{2}=x$

$cosx=\frac{\sqrt{3}}{2}$

$x=\frac{\pi }{6}$ where $x\in [0,\pi ]$

$2co{s}^{-1}\frac{\sqrt{3}}{2}=2\frac{\pi }{6}=\frac{\pi }{3}$

Then the expression reduces to $ta{n}^{-1}\left(2sin\frac{\pi }{3}\right)$

Now we are going to reduce the term $2sin\frac{\pi }{3}$

We know that $sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}$

$2sin\frac{\pi }{3}=2\frac{\sqrt{3}}{2}=\sqrt{3}$

The expression becomes, $ta{n}^{-1}\left(\sqrt{3}\right)$

We know that $ta{n}^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$

Therefore, $ta{n}^{-1}\left[2sin\left(2co{s}^{-1}\frac{\sqrt{3}}{2}\right)\right]=\frac{\pi }{3}$.