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

Solution:

(i) LHS $=2\left(\sin \frac{5 \pi}{12}\right)\left(\sin \frac{\pi}{12}\right)$

$=\cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)-\cos \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right) \quad[\because 2 \sin A \sin B=\cos (A-B)-\cos (A+B)]$

$=\cos \frac{\pi}{3}-\cos \frac{\pi}{2}$

$=\frac{1}{2}-0$

$=\frac{1}{2}$

RHS $=\frac{1}{2}$

Hence, LHS $=$ RHS

(ii) $\mathrm{LHS}=2\left(\cos \frac{5 \pi}{12}\right)\left(\cos \frac{\pi}{12}\right)$

$=\cos \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)+\cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right) \quad[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$

$=\cos \frac{\pi}{2}+\cos \frac{\pi}{3}$

$=0+\frac{1}{2}$

$=\frac{1}{2}$

RHS $=\frac{1}{2}$

Hence, LHS $=$ RHS

(iii) $\mathrm{LHS}=2\left(\sin \frac{5 \pi}{12}\right)\left(\cos \frac{\pi}{12}\right)$

$=\sin \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)+\sin \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right) \quad[\because 2 \sin A \cos B=\sin (A+B)+\sin (A-B)]$

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

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

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

RHS $=\frac{2+\sqrt{3}}{2}$ Hence, LHS $=$ RHS