# How do you prove cos(x + pi/6) + sin(x + pi/3) = (sqrt 3)cos x?

Mar 19, 2016

We use some basics like
$C o s \left(90 - A\right) = \sin A$

$\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}$

#### Explanation:

let me rewrite the given LHS;

$\sin \left(x + \frac{\pi}{3}\right) + \cos \left(x + \frac{x}{6}\right)$

$\sin \left(x + \frac{\pi}{3}\right) + \sin \left(\frac{\pi}{2} - \left(x + \frac{\pi}{6}\right)\right)$

$\sin \left(x + \frac{\pi}{3}\right) + \sin \left(\frac{\pi}{2} - x - \frac{\pi}{6}\right)$

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

Now recall

$\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}$

we get

$= 2 \sin \left(\frac{\left(x + \frac{\pi}{3}\right) + \left(\frac{\pi}{3} - x\right)}{2}\right) \cos \left(\frac{\left(x + \frac{\pi}{3}\right) - \left(\frac{\pi}{3} - x\right)}{2}\right)$

$= 2 \sin \left(\frac{\pi}{3}\right) \cos x$

$= 2 \cdot \frac{\sqrt{3}}{2} \cdot \cos x$

$= \cancel{2} \cdot \frac{\sqrt{3}}{\cancel{2}} \cdot \cos x$

$= \sqrt{3} \cos x = R H S$