Question

Prove the identity of cos(x-pi/3)+sin(pi/6-x)=cosx

Answer 1

Answer:

$\sf \cos \left( x - \dfrac{\pi}{3} \right) + \sin \left(\dfrac{\pi}{6} - x \right) = \cos x \ [Proved]$

Step-by-step explanation:

Given :

$\sf \cos \left( x - \dfrac{\pi}{3} \right) + \sin \left(\dfrac{\pi}{6} - x \right) = \cos x$

$\sf L.H.S = \cos \left( x - \dfrac{\pi}{3} \right) + \sin \left(\dfrac{\pi}{6} - x \right)$

Converting sin into cos by using complementary formula :

$\sf \rightarrow \cos \left( x - \dfrac{\pi}{3} \right) + \cos \left(\dfrac{\pi}{2}- \left( \dfrac{\pi}{6} - x \right) \right)$

$\sf \rightarrow \cos \left( x - \dfrac{\pi}{3} \right) + \cos \left(\dfrac{\pi}{3}+ x \right)$

Now using formula cos C + cos D  :

$\sf \cos \ C + \cos \ D = 2 \cos \left( \dfrac{C+D}{2} \right).cos \left( \dfrac{C-D}{2} \right)$

$\sf \rightarrow \cos \left(\dfrac{\pi}{3}+ x \right)+ \cos \left( x - \dfrac{\pi}{3} \right)$

$\sf \rightarrow 2 \cos \left(\dfrac{\dfrac{\pi}{3}+ x + x - \dfrac{\pi}{3}}{2} \right).\cos \left(\dfrac{\dfrac{\pi}{3}+ x - x + \dfrac{\pi}{3}}{2}\right)$

$\sf \rightarrow 2 \cos \left(\dfrac{2.x}{2} \right).\cos \left(\dfrac{2.\dfrac{\pi}{3}}{2}\right)$

$\sf \rightarrow 2 \cos \left(x \right).\cos \left(\dfrac{\pi}{3} \right)$

We know :

$\cos \left(\dfrac{\pi}{3} \right) = \dfrac{1}{2}$

$\sf \rightarrow 2 \cos \left(x \right).\left(\dfrac{1}{2} \right)$

$\sf \rightarrow \cos \left(x \right)$

Since L.H.S. = R.H.S.

Hence proved.