Question

How do you find the value of $\sin \left(\frac{5\pi }{12}\right)$?

Answer 1

Determine the value of $\sin \left(\frac{5\pi }{12}\right)$:

The expression $\left(\frac{5\pi }{12}\right)$ can be written as $\frac{\pi }{4}+\frac{\pi }{6}$.

The trigonometric identity for sum of angles for the sum of angles is $\sin \left(A+B\right)=\sin A\cos B+\sin B\cos A$.

Use the identity to find the value of $\sin \left(\frac{5\pi }{12}\right)$.

$\begin{array}{rcl}\sin \left(\frac{\pi }{4}+\frac{\pi }{6}\right)& =& \sin \frac{\pi }{4}\cos \frac{\pi }{6}+\sin \frac{\pi }{6}\cos \frac{\pi }{4}\\ & \Rightarrow & \sin \left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{\sqrt{2}}\\ & \Rightarrow & \sin \left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\frac{1+\sqrt{3}}{2\sqrt{2}}\end{array}$

Hence, the required value is $\begin{array}{l}\frac{1+\sqrt{3}}{2\sqrt{2}}\end{array}$.