How do you prove sin[x+(pi/6)] + sin[x-(pi/6)] = (sqrt3)/2?

Nov 12, 2015

$\sin \left(x + \frac{\pi}{6}\right) + \sin \left(x - \frac{\pi}{6}\right) = \sqrt{3} \sin \left(x\right)$

Explanation:

Using the trigonometric identity
$\sin \left(\alpha + \beta\right) = \sin \left(\alpha\right) \cos \left(\beta\right) + \cos \left(\alpha\right) \sin \left(\beta\right)$

$\sin \left(x + \frac{\pi}{6}\right) = \sin \left(x\right) \cos \left(\frac{\pi}{6}\right) + \cos \left(x\right) \sin \left(\frac{\pi}{6}\right)$
and
$\sin \left(x - \frac{\pi}{6}\right) = \sin \left(x\right) \cos \left(- \frac{\pi}{6}\right) + \cos \left(x\right) \sin \left(- \frac{\pi}{6}\right)$

As the sine function is odd ($\sin \left(- x\right) = - \sin \left(x\right)$) and the cosine function is even ($\cos \left(- x\right) = \cos \left(x\right)$), we get

$\sin \left(x - \frac{\pi}{6}\right) = \sin \left(x\right) \cos \left(\frac{\pi}{6}\right) - \cos \left(x\right) \sin \left(\frac{\pi}{6}\right)$

$\sin \left(x + \frac{\pi}{6}\right) + \sin \left(x - \frac{\pi}{6}\right) = 2 \sin \left(x\right) \cos \left(\frac{\pi}{6}\right)$

As $\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ we get the result

$\sin \left(x + \frac{\pi}{6}\right) + \sin \left(x - \frac{\pi}{6}\right) = \sqrt{3} \sin \left(x\right)$