Question

prove that cosx + cos(2pi/3+x) + cos (2pi/3-x)=0

Answer 1

L.H.S

$cos x + cos( \frac{2 \pi }{3} + x)+cos( \frac{2\pi}{3} -x)$
Using cos(A+B) = cosAcosB-sinAsinB and cos(A-B) = cosAcosB+sinAsinB we get,
[tex]cosx + cos \frac{2\pi}{3}cosx-sin \frac{2\pi}{3}sinx + cos \frac{2\pi}{3}cosx +sin \frac{2\pi}{3}sinx \\ cosx + cos \frac{2\pi}{3}cosx+cos \frac{2\pi}{3}cosx [/tex]
As $cos \frac{2\pi}{3}=- \frac{1}{2}$
[tex]cosx(1+cos \frac{2\pi}{3}+cos \frac{2\pi}{3} ) \\ cosx(1- \frac{1}{2}- \frac{1}{2} ) \\ cosx(0) \\ 0[/tex]
Thus L.H.S = R.H.S Proved

Answer 2

We know that cos(A+B)+cos(A-B)=2cosAcosB So here cosx+cos(120+x)+cos(120-x)=cosx+2cos120cosx =cosx+2(-1/2)cosx..,...(cos120=-1/2) =cosx-cosx =0 Hence proved...