Question

If $0\le \alpha <\beta <\gamma \le 2\pi$ and if $f\left(x\right)=\cos (x+\alpha )+\cos (x+\beta )+\cos (x+\gamma )=0\forall z\in R$, then value of $\gamma -\alpha$ is equal to

• A. $\frac{2\pi }{3}$
• B. $\frac{4\pi }{3}$
• C. $\frac{\pi }{3}$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{2\pi }{3}$

solve:

Given,

$f\left(x\right)=\cos (x+\alpha )+\cos (x+\beta )+\cos (x+\gamma )=0$

we know that $\cos (x+y)=\cos x\cos y-\sin x\sin y$

$\Rightarrow \cos x\cos \alpha -\sin x\sin \alpha +\cos x\cos \beta -\sin x\sin \beta$

$+\cos x\cos \gamma -\sin x\sin \gamma =0$

on seprating the terms of cosx and sinx we get,

$\cos x(\cos \alpha +\cos \beta +\cos \gamma )-\sin x(\sin \alpha +\sin \beta +\sin \gamma )=0$

both sinx and cosx simultaneousty can not be zero

$\Rightarrow \cos \alpha +\cos B+\cos \gamma =\sin \alpha +\sin B+\sin \gamma =0$

$\Rightarrow \cos \alpha +\cos \beta +\cos \gamma =0$

$=>\cos \alpha +\cos \gamma =-\cos B-\left(1\right)$

and $\sin \alpha +\sin B+\sin \gamma =0$

$\Rightarrow \sin \alpha +\sin \gamma =-\sin B-\left(ii\right)$

on Squaring and adding equation $\left(i\right)$ and

(ii) we get,

${\cos }^2\alpha +{\cos }^2\gamma +2\cos \alpha \cos \gamma +{\sin }^2\alpha +{\sin }^2\gamma +2\sin \alpha \sin \gamma ={\cos }^2\beta +{\sin }^{2}B$

$\Rightarrow 2+2\cos (\alpha -r)=1$

$\Rightarrow \cos (\alpha -r)=\frac{-1}{2}$

$\Rightarrow \cos (\gamma -\alpha )=-\frac{1}{2}$

$\Rightarrow \gamma -\alpha =\frac{2\pi }{3}$