Question

What is the value of $\cos 15°$?

Answer 1

Find the value of $\cos 15°$.

We know that,

$\begin{array}{rcl}\cos \left(15°\right)& =& \cos \left(45°-30°\right)\\ & =& \cos 45°\cos 30°+\sin 45°\sin 30°\left\{\because \cos \left(A-B\right)=\cos A\cos B+\sin A\sin B\right\}\\ & =& \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\left\{\mathrm{Put}\cos 45°=\frac{1}{\sqrt{2}},\cos 30°=\frac{\sqrt{3}}{2},\sin 45°=\frac{1}{\sqrt{2}}\sin 30°=\frac{1}{2}\right\}\\ & =& \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\\ & =& \frac{\sqrt{3}+1}{2\sqrt{2}}\end{array}$

Hence, the required value is $\cos \left(15°\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}$