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Q.
Find the value of $sin \frac{5\pi}{12}\,sin \, \frac{\pi}{12}$.
Trigonometric Functions
Solution:
We have, $sin \frac{5\pi}{12}\cdot sin \frac{\pi}{12}=\frac{1}{2}\left[2\,sin \frac{5\pi}{12}\cdot sin \frac{\pi}{12}\right]$
$=\frac{1}{2}\left[cos\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)-cos\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)\right]$
$=\frac{1}{2}\left(cos \frac{\pi}{3}-cos \frac{\pi}{2}\right)=\frac{1}{2}\left[\frac{1}{2}-0\right]$
$=\frac{1}{4}$