Question

If $\cos \alpha +\cos \beta +\cos \gamma =\sin \alpha +\sin \beta +\sin \gamma =0,$ then $\cos (2\alpha -\beta -\gamma )+\cos (2\beta -\gamma -\alpha )+\cos (2\gamma -\alpha -\beta )=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Let $x=\cos \alpha +i\sin \alpha ,y=\cos \beta +i\sin \beta ,z=\cos \gamma +i\sin \gamma$
$\Rightarrow x+y+z=(\cos \alpha +\cos \beta +\cos \gamma )+i(\sin \alpha +\sin \beta +\sin \gamma )$
$\Rightarrow x+y+z=0$
$\Rightarrow x^3+y^3+z^3=3xyz$

$\frac{x^3+y^3+z^3}{xyz}=3\Rightarrow \frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3$
$\Rightarrow \frac{e^{2i\alpha }}{e^{i\beta }{e}^{i\gamma }}+\frac{e^{2i\beta }}{e^{i\gamma }{e}^{i\alpha }}+\frac{e^{2i\gamma }}{e^{i\alpha }{e}^{i\beta }}=3$
$\Rightarrow e^{i(2\alpha -\beta -\gamma )}+e^{i(2\beta -\gamma -\alpha )}+e^{i(2\gamma -\alpha -\beta )}=3$
Comparing real part on both sides
$\cos (2\alpha -\beta -\gamma )+\cos (2\beta -\gamma -\alpha )+\cos (2\gamma -\alpha -\beta )=3$