# How do you find the exact value of cos62 using the sum and difference, double angle or half angle formulas?

Jan 8, 2018

$\cos \left({62}^{\circ}\right) = \cos \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$

#### Explanation:

Best way to start is to convert the angle measure from degrees to radians. A quick refresher on how to do this;

Radians = Degrees $\times \frac{\pi}{180}$

Radians = $62 \times \frac{\pi}{180} = \frac{62 \pi}{180} = \frac{7 \pi}{12}$

$\therefore \cos \left({62}^{\circ}\right) = \cos \left(\frac{7 \pi}{12}\right)$

$\cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right)$

The trigonometric identity that we are going to use is the angle sum identity for cosine;

$\textcolor{red}{\cos \left(A + B\right) = \cos \left(A\right) \cos \left(B\right) - \sin \left(A\right) \sin \left(B\right)}$

Entering in the appropriate angles;

$\cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$

Evaluating the trigonometric ratios yields;

$\cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \left(\frac{1}{2}\right) \left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)$

Hence,

$\cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$