# How do you use the half angle identity to find exact value of sin (-17pi/12)+sin(pi/12)?

Sep 7, 2015

Find:$\sin \left(\frac{- 17 \pi}{12}\right) + \sin \left(\frac{\pi}{12}\right)$

Ans: zero

#### Explanation:

Call $\sin \left(\frac{- 17 \pi}{12}\right) = \sin t$
$\sin \left(\frac{- 17 \pi}{12}\right) = \sin t = \sin \left(\frac{- 5 \pi}{12} - 2 \pi\right) = \sin \left(\frac{- 5 \pi}{12}\right) = - \sin \left(\frac{5 \pi}{12}\right)$
$\cos 2 t = \cos \left(\frac{10 \pi}{12}\right) = \cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2}$
Apply the trig identity: $\cos 2 a = 2 {\cos}^{2} a - 1$
$\cos \left(\frac{5 \pi}{6}\right) = \cos 2 t = - \frac{\sqrt{3}}{2} = 1 - 2 {\sin}^{2} t$
$2 {\sin}^{2} t = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$
${\sin}^{2} t = \frac{2 + \sqrt{3}}{4}$
$\sin \left(\frac{- 17 \pi}{12}\right) = \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$
Since the arc (-17pi/12) is located in Quadrant III, its sin is positive, then $\sin \left(\frac{- 17 \pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$
Call $\sin \left(\frac{\pi}{12}\right) = \sin u$
$\cos 2 u = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
$\cos 2 u = \frac{\sqrt{3}}{2} = 1 - 2 {\sin}^{2} u$
${\sin}^{2} u = \frac{2 - \sqrt{3}}{4}$
$\sin \left(\frac{\pi}{12}\right) = \sin u = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$
Since $\sin \frac{\pi}{12} > 0$, then
$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ (2)
From (1) and (2), we get: $\sin \left(\frac{- 17 \pi}{12}\right) + \sin \left(\frac{\pi}{12}\right) =$
$\frac{\sqrt{2 + \sqrt{3}}}{2} + \frac{\sqrt{2 - \sqrt{3}}}{2} = 1$