# How do you verify the identity sin(pi/6+x)+sin(pi/6-x)=cosx?

Feb 3, 2017

see explanation below

#### Explanation:

expand $\sin \left(A + B\right) = \sin A \cos A + \sin B \cos A$ and,
$\sin \left(A - B\right) = \sin A \cos A - \sin B \cos A$

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

$= 2 \sin \left(\frac{\pi}{6}\right) \cos x$, where $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

therefore,

$= 2 \sin \left(\frac{\pi}{6}\right) \cos x = 2 \left(\frac{1}{2}\right) \cos x = \cos x$,

Feb 3, 2017

Explained below.

#### Explanation:

Expand the left hand side,

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

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

=$2 \left(\frac{1}{2}\right) \cos x$

= cos x = right hand side