Question

If $\cos x+\cos y+\cos z=0$ and $\sin x+\sin y+\sin z=0$, show that,
$\cos (x-y)=\cos (y-z)=\cos (z-x)=-\frac{1}{2}.$

Answer 1

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

$\sin x+\sin y+\sin z=0$

Show that $\cos (x-y)=\cos (y-z)=\cos (z-x)=\frac{-3}{2}$

$\because \cos x+\cos y=-\cos z$

or, ${\cos }^{2}x+{\cos }^{2}y+2\cos x.\cos y={\cos }^{2}z$

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

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

Adding both side we get,

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

or, $1+1+2\cos (x-y)=1$

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

Similarly we can prove,

$\sin x+\sin z=\sin y$

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

$⟹\cos x+\cos z=-\cos y$

$⟹{\cos }^{2}x+{\cos }^{2}z+2\cos x.\cos z={\cos }^{2}y$

adding both we get,

$1+1+2\cos (z-x)=1$

$\cos (z-x)=\frac{-1}{2}$

similarly $\cos (y-z)=\frac{-1}{2}$