Question

Prove that cos15°-sin15°=1/√2

Answer 1

Hope this helps you in a positive way.

Answer 2

$\begin{array}{c}{\cos 15^{\circ}-\sin 15^{\circ}=\frac{1}{\sqrt{2}}} \\ {\quad \cos 15^{\circ}=\cos \left(45^{\circ}-30^{\circ}\right)}\end{array}$

$\begin{array}{c}{\sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)} \\ {\cos (A-B)=\cos A \times \cos B+\sin A \times \sin B}\end{array}$

$\begin{array}{c}{\sin (A-B)=\sin A \times \cos B-\cos A \times \sin B} \\ {\cos \left(45^{\circ}-30^{\circ}\right)=\cos 45^{\circ} \times \cos 30^{\circ}+\sin 45^{\circ} \times \sin 30^{\circ}} \\ {\cos 15^{\circ}=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\frac{\sqrt{3}+1}{2 \sqrt{2}}} \\ {\sin \left(45^{\circ}-30^{\circ}\right)=\sin 45^{\circ} \times \cos 30^{\circ}-\cos 45^{\circ} \times \sin 30^{\circ}}\end{array}$  

$\begin{array}{c}{\sin 15^{\circ}=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}} \\ {\cos 15^{\circ}-\sin 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}}} \\ {\quad=\frac{3-1}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}} \\ \bold{{\cos 15^{\circ}-\sin 15^{\circ}=\frac{1}{\sqrt{2}}}\end{array}}$

∴Hence Proved.