Question

Evaluate $2\sin \left(\frac{\pi }{12}\right)$

Answer 1

We have, $2\sin \left(\frac{\pi }{12}\right)$

$=2\sin (\frac{\pi }{3}-\frac{\pi }{4})$ [$\because \frac{\pi }{3}-\frac{\pi }{4}=\frac{4\pi -3\pi }{12}=\frac{\pi }{12}$]

$=2\left[\sin \right(\frac{\pi }{3}\left)\cos \right(\frac{\pi }{4})-\cos (\frac{\pi }{3}\left)\sin \right(\frac{\pi }{4}\left)\right]$ [$\because \sin (A-B)=\sin A\times \cos B-\cos A\times \sin B$]

$=2\left[\right(\frac{\sqrt{3}}{2}\left)\right(\frac{1}{\sqrt{2}})-(\frac{1}{2}\left)\right(\frac{1}{\sqrt{2}})$] [$\because \sin \left(\frac{\pi }{3}\right)=\left(\frac{\sqrt{3}}{2}\right),\cos \left(\frac{\pi }{4}\right)=\left(\frac{1}{\sqrt{2}}\right),\cos \left(\frac{\pi }{3}\right)=\left(\frac{1}{2}\right),\sin \left(\frac{\pi }{4}\right)=\left(\frac{1}{\sqrt{2}}\right)$ ]

$=2\left[\right(\frac{\sqrt{3}-1}{2\sqrt{2}}]$

$=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)$