How do you use the half angle identify to find the exact value of cos((7pi)/12)?

Nov 11, 2016
• sqrt(2 + sqrt3)/2

Explanation:

Using trig unit circle and trig table -->
$\cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{12} + \pi\right) = - \cos \left(\frac{\pi}{12}\right)$.
Find cos (pi/12) by using trig identity:
$2 {\cos}^{2} a = 1 + \cos 2 a$
$2 {\cos}^{2} \left(\frac{\pi}{12}\right) = 1 + \cos \left(\frac{\pi}{6}\right) = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$
${\cos}^{2} \left(\frac{\pi}{12}\right) = \frac{2 + \sqrt{3}}{4}$
$\cos \left(\frac{\pi}{12}\right) = \pm \frac{\sqrt{2 + \sqrt{3}}}{2.}$
Since $\cos \left(\frac{\pi}{12}\right)$ is positive, only the positive value is accepted.
Finally,
$\cos \left(\frac{7 \pi}{12}\right) = - \cos \left(\frac{\pi}{12}\right) = - \frac{\sqrt{2 + \sqrt{3}}}{2}$
Check by calculator.
$\cos \left(\frac{7 \pi}{12}\right) = - \cos \left(\frac{\pi}{12}\right) = - \cos 15 = - 0.97$
$- \frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{3.73}}{2} = \frac{1.93}{2} = 0.97$. OK