Question

Given $\cos (A-B)=\cos A\cos B+\sin A\sin B$. Taking suitable $A$ and $B$, find $\cos {15}^{\circ }$.

Answer 1

We have,

$\cos (A-B)=cosAcosB+sinAsinB$ …… (1)

Then, $\cos {15}^o=?$

Let $A={45}^o$ and $B={30}^o$

Put the value of $A$ and $B$ in equation (1) and we get,

$\cos ({45}^o-{30}^o)=\cos {45}^o\cos {30}^o+\sin {45}^o\sin {30}^o$

$\cos {15}^o=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$

$\cos {15}^o=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Hence, this is the answer.