# How do you prove sin(x+[pi/6])*sin(x-[pi/6]) = (sin x)^2 - 1/4?

May 4, 2016

see below

#### Explanation:

Use Properties: $\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$

$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$

Left Side:$= \left(\sin x \cos \left(\frac{\pi}{6}\right) + \cos x \sin \left(\frac{\pi}{6}\right)\right) \times \left(\sin x \cos \left(\frac{\pi}{6}\right) - \cos x \sin \left(\frac{\pi}{6}\right)\right)$

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

$= {\sin}^{2} x \cdot {\left(\frac{\sqrt{3}}{2}\right)}^{2} - {\cos}^{2} x \cdot {\left(\frac{1}{2}\right)}^{2}$

$= \frac{3}{4} {\sin}^{2} x - \frac{1}{4} {\cos}^{2} x$

$= \frac{3}{4} {\sin}^{2} x - \frac{1}{4} \left(1 - {\sin}^{2} x\right)$

$= \frac{3}{4} {\sin}^{2} x - \frac{1}{4} + \frac{1}{4} {\sin}^{2} x$

$= \left(\frac{3}{4} {\sin}^{2} x + \frac{1}{4} {\sin}^{2} x\right) - \frac{1}{4}$

$= {\sin}^{2} x - \frac{1}{4}$

$=$ Right Side