# Find the value of cos11π/12?

Jun 12, 2018

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

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

•color(white)(x)cos(x+y)=cosxcosy-sinxsiny

$\text{note that } \frac{11 \pi}{12} = \frac{2 \pi}{3} + \frac{\pi}{4}$

$\cos \left(\frac{11 \pi}{12}\right) = \cos \left(\frac{2 \pi}{3} + \frac{\pi}{4}\right)$

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

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

$= \left(- \frac{1}{2} \times \frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\right)$

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

Jun 12, 2018

$- \frac{\sqrt{2 + \sqrt{3}}}{2}$

#### Explanation:

$\cos \left(\frac{11 \pi}{12}\right) = \cos \left(- \frac{\pi}{12} + \frac{12 \pi}{12}\right) = \cos \left(- \frac{\pi}{12} + \pi\right) =$
$= - \cos \left(- \frac{\pi}{12}\right) = - \cos \left(\frac{\pi}{12}\right)$
Find $\cos \left(\frac{\pi}{12}\right)$ by using trig identity:
$2 {\cos}^{2} a = 1 + \cos 2 a$.
In this case:
$2 {\cos}^{2} \left(\frac{\pi}{12}\right) = 1 + \cos \left(\frac{\pi}{6}\right) = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$
${\cos}^{2} \left(\frac{\pi}{12}\right) = \frac{2 + \sqrt{3}}{4}$
$\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$ (because $\cos \left(\frac{\pi}{12}\right)$ is positive)
Finally,
$\cos \left(\frac{11 \pi}{12}\right) = - \cos \left(\frac{\pi}{12}\right) = - \frac{\sqrt{2 + \sqrt{3}}}{2}$
Check by calculator.
$\cos \left(\frac{11 \pi}{12}\right) = \cos {165}^{\circ} = - 0.966$
- $\frac{\sqrt{2 + \sqrt{3}}}{2} = - \frac{1.932}{2} = - 0.966$. Proved