Question

If $\cos x+\cos y+\cos z=0$ $=\sin x+\sin y+\sin z$, then possible value of $\cos (x-y)$ is

Answer 1

Let $\cos x+\cos y+\cos z=0$

or, $\cos x+\cos y=-\cos z$

Squaring both sides we get,

${\cos }^{2}x+2\cos x.\cos y+{\cos }^y={\cos }^{2}z$.........(1).
Again let $\sin x+\sin y+\sin z=0$

or, $\sin x+\sin y=-\sin z$

Squaring both sides we get,

${\sin }^{2}x+2\sin x.\sin y+{\sin }^{2}y={\sin }^{2}z$.........(2).

Now adding (1) and (2) we get,

$2+2\left(\cos \right(x-y\left)\right)=1$

or, $\cos (x-y)=-\frac{1}{2}$