# How do you find the exact value for cos165 using the half‐angle identity?

Feb 13, 2015

The answer is: $- \frac{\sqrt{6} + \sqrt{2}}{4}$

The formula of alf‐angle is:

$\cos \left(\frac{\alpha}{2}\right) = \pm \sqrt{\frac{1 + \cos \alpha}{2}}$, the $\pm$ in this case becomes $-$ because the angle of 165° is in the second quadrant and there the cosine is negative.

So:

cos165°=-sqrt((1+cos330°))/2=-sqrt((1+sqrt3/2)/2)=-sqrt((2+sqrt3)/4)=-sqrt(2+sqrt3)/2

or, using the formula of double radical, that says:

$\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{c}}{2}} \pm \sqrt{\frac{a - \sqrt{c}}{2}}$, that is useful when $c = {a}^{2} - b$ is a square.

So: $c = 4 - 3 = 1$, and than:

$- \frac{\sqrt{2 + \sqrt{3}}}{2} = - \frac{1}{2} \left[\sqrt{\frac{2 + 1}{2}} + \frac{\sqrt{2 - 1}}{2}\right] = - \frac{1}{2} \left[\sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}\right] =$

$= - \frac{1}{2} \left(\frac{\sqrt{3}}{\sqrt{2}} + \frac{\sqrt{1}}{\sqrt{2}}\right) = - \frac{1}{2} \left(\frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\right) = - \frac{1}{2} \left(\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}\right) = - \frac{\sqrt{6} + \sqrt{2}}{4}$