Question

cosx+cos(120'-x)+cos(120'+x)=0​

Answer 1

Answer:

0

Step-by-step explanation:

$cosx + cos (120-x) + cos (120+x)=>. cosx + 2cos [(120-x+120+x) /2] cos [(120-x-120-x) /2]=>. cosx + 2cos120 cos(-x)=>. cosx + 2cos(180-60) cosx=>. cosx + 2 (-cos 60) cosx=>. cosx - 2 × 1/2 cosx=>. cosx - cosx=>. 0 hence proved$

cosx + cos (120-x) + cos (120+x)

=>. cosx + 2cos [(120-x+120+x) /2] cos [(120-x-120-x) /2]

=>. cosx + 2cos120 cos(-x)

=>. cosx + 2cos(180-60) cosx

=>. cosx + 2 (-cos 60) cosx

=>. cosx - 2 × 1/2 cosx

=>. cosx - cosx

=>. 0 hence proved

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Answer 2

Step-by-step explanation:

LHS:- cos x+ cos (120°-x)+cos(120°+x)

cos x+ 2× cos 120°×cos x

cos (a+b)+ cos (a-b)=2cosAcosB

cos x +2cos(180°-60°) cos x

cos{1+2(-cos 60°)}

cos {1+2(1/2)}

cos x (1-1)

0=RHS

PROVED