How do you find the exact values of cos(11pi/12) using the half angle formula?

Aug 27, 2015

Find $\cos \left(\frac{11 \pi}{12}\right)$

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

Explanation:

Call $\cos \left(\frac{11 \pi}{12}\right) = \cos t$
$\cos 2 t = \cos \left(\frac{22 \pi}{12}\right) = \cos \left(\frac{11 \pi}{6}\right) = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
Apply the trig identity: $\cos 2 t = 2 {\cos}^{2} t - 1$
$\cos 2 t = \frac{\sqrt{3}}{2} = 2 {\cos}^{2} t - 1$
$2 {\cos}^{2} t = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$
${\cos}^{2} t = \frac{2 + \sqrt{3}}{4}$
$\cos t = \cos \left(\frac{11 \pi}{12}\right) = \pm \frac{\sqrt{2 + \sqrt{3}}}{2.}$
Since the arc ((11pi)/12) is located in Quadrant II, only the negative answer is accepted.
$\cos \left(\frac{11 \pi}{12}\right) = - \frac{\sqrt{2 + \sqrt{3}}}{2}$

Check by calculator.
$A r c \left(\frac{11 \pi}{12}\right) = 165$ deg-> $\cos \left(\frac{11 \pi}{12}\right) = \cos 165 = - 0.97 .$
$- \left(\frac{\sqrt{2 + \sqrt{3}}}{2}\right) = - 0.97$. OK