Question

Solve it :-
$\begin{array}{c}\cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}\\ \sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}\end{array}$ $\left(why?\right)$

Answer 1

$\left(i\right)\cos {15}^{\circ }=\cos ({45}^{\circ }-{30}^{\circ })$

$=\cos {45}^{\circ }\cos {30}^{\circ }+\sin {45}^{\circ }\sin {30}^{\circ }$ using $\cos (A-B)=\cos A\cos B+\sin A\sin B$

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

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

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

$\left(ii\right)\sin {15}^{\circ }=\sin ({45}^{\circ }-{30}^{\circ })$

$=\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ }$ using $\sin (A-B)=\sin A\cos B+\cos A\sin B$

$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$

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

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