Question

Evaluate $\cos {15}^0$ using the for formula :$\cos (A-B)=\cos A\cos B+\sin A\sin B$

Answer 1

Solution:-

$\cos {15}^0=\cos ({45}^0-{30}^0)$

$\Rightarrow =\cos {45}^0\cos {30}^0+\sin {45}^0\sin {30}^0$

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

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

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

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

$\Rightarrow =\frac{(\sqrt{3}+1)\times \sqrt{2}}{2{\left(\sqrt{2}\right)}^2}$

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

$\Rightarrow =\frac{\sqrt{6}+\sqrt{2}}{4}$

Hence, $\cos 15=\frac{\sqrt{6}+\sqrt{2}}{4}$.