Prove that,4 sin15° sin75°=$\sqrt{2}$ (cos 105° + sin75°)

Prove that,4 sin15° sin75°=$\sqrt{2}$ (cos 105° + sin75°)

L.H.S:

4 sin15° sin75°

= 2 (2 sin15° sin75°)

= 2{cos(15°+75°)+cos(15°-75°)}

=2 {cos90° + cos(-60°)}

=2 (cos90° + cos60°)

= 2 $\left(0 + \frac{1}{2}\right)$

= 2×$\frac{1}{2}$

= 1

R.H.S:

$\sqrt{2}$ (cos 105° + sin75°)

=  $\sqrt{2}$ {cos105°+ sin(90°-15°)}

= $\sqrt{2}$ (cos105°+cos15°)

= $\sqrt{2}$  ×2 cos $\frac{105° + 15°}{2}$ cos $\frac{105° – 15°}{2}$

= 2 $\sqrt{2}$  cos 60° cos45°

= 2$\sqrt{2}$ × $\frac{1}{2}$× $\frac{1}{\sqrt{2}}$

= 1

∴L.H.S = R.H.S [Proved]