How do you find the value of sin -((5pi)/12) using the double or half angle formula?

Oct 25, 2016

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

Explanation:

Unit circle and property of complementary arcs -->
$\sin \left(\frac{- 5 \pi}{12}\right) = \sin \left(\frac{\pi}{12} - \frac{6 \pi}{12}\right) = \sin \left(\frac{\pi}{12} - \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{12}\right)$
Evaluate $\cos \left(\frac{\pi}{12}\right)$, knowing $\cos \left(\frac{2 \pi}{12}\right) = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
Use trig identity
$2 {\cos}^{2} a = 1 + \cos 2 a$
$2 {\cos}^{2} \left(\frac{\pi}{12}\right) = 1 + \cos \left(\frac{\pi}{6}\right) = 1 + s q t \frac{3}{2} = \frac{2 + \sqrt{3}}{2}$
${\cos}^{2} \left(\frac{\pi}{12}\right) = \frac{2 + \sqrt{3}}{4}$
$\cos \left(\frac{\pi}{12}\right) = \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$
Since cos ((pi/12) is positive, then only the positive value is accepted.
Finally:
$\sin \left(\frac{- 5 \pi}{12}\right) = \cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$