# Prove that

Question:

Prove that

(i) $2 \sin \frac{5 \pi}{12} \sin \frac{\pi}{12}=\frac{1}{2}$

(ii) $2 \cos \frac{5 \pi}{12} \cos \frac{\pi}{12}=\frac{1}{2}$

(iii) $2 \sin \frac{5 \pi}{12} \cos \frac{\pi}{12}=\frac{(2+\sqrt{3)}}{2}$

Solution:

(i) $2 \sin \frac{5 \pi}{12} \cdot \sin \frac{\pi}{12}=-\left(\cos \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)-\cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)\right)$

………[Using –2sinx.siny = cos(x + y)–cos (x–y)]

$=-\left(\cos \frac{6 \pi}{12}-\cos \frac{4 \pi}{12}\right)$

$=-\left(\cos \frac{\pi}{2}-\cos \frac{\pi}{3}\right) \Rightarrow-\left(0-\frac{1}{2}\right)=\frac{1}{2}$

(ii) $2 \cos \frac{5 \pi}{12} \cdot \cos \frac{\pi}{12}=\cos \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)+\cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)$

................$.[$ using $2 \cos x . \cos y=\cos (x+y)+\cos (x-y)]$

$=\cos \frac{6 \pi}{12}+\cos \frac{4 \pi}{12} \Rightarrow \cos \frac{\pi}{2}+\cos \frac{\pi}{3}=0+\frac{1}{2}$

$=\frac{1}{2}$

(iii) $2 \sin \frac{5 \pi}{12} \cdot \cos \frac{\pi}{12}=\sin \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)+\sin \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)$

$\ldots[U \sin g 2 \sin x \cdot \cos y=\sin (x+y)+\sin (x-y)]$

$=\sin \frac{6 \pi}{12}+\sin \frac{4 \pi}{12} \Rightarrow \sin \frac{\pi}{2}+\sin \frac{\pi}{3}$

$=1+\frac{\sqrt{3}}{2} \Rightarrow \frac{2+\sqrt{3}}{2}$