Question:
Prove that
$2 \sin 75^{\circ} \sin 15^{0}=\frac{1}{2}$
Solution:
L.H.S
$=2 \sin 75^{\circ} \sin 15^{\circ}$
$=2 \sin \left(45^{\circ}+30^{\circ}\right) \sin \left(45^{\circ}-30^{\circ}\right)$
$=\cos \left(45^{\circ}-30^{\circ}-45^{\circ}-30^{\circ}\right)-\cos \left(45^{\circ}+30^{\circ}+45^{\circ}-30^{\circ}\right)$
$=\cos \left(-60^{\circ}\right)-\cos 90^{\circ}$
$=\cos 60^{\circ}-0$
$=\frac{1}{2}$