Question:
Prove that
$\cos 15^{\circ}-\sin 15=\frac{1}{\sqrt{2}}$
Solution:
L.H.S
$\Rightarrow \cos 15^{\circ}-\sin 15^{\circ}$
$\Rightarrow \cos \left(45^{\circ}-30^{\circ}\right)-\sin \left(45^{\circ}-30^{\circ}\right)$
$\Rightarrow\left(\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}\right)-\left(\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}\right)$
$\Rightarrow\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)-\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)$
$\Rightarrow \frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}-\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}$
$\Rightarrow \frac{2}{2 \sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}}$