Question

$Cos(x-y)+Cos(y-z)+Cos(z-x)=\frac{-3}{5}$
Find (1) $Cosx+Cosy+Cosz=?$

(2) $Sinx+Siny+Sinz=?$

Answer 1

If $\cos (x-y)+\cos (y-z)+\cos (z-x)=-\frac{3}{2}$

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

$\cos x+\cos y+\cos z=?$

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

$\Rightarrow \cos x\cos y+\sin x\sin y+\cos y\cos z+\sin y\sin z+\cos z\cos x+\sin z\sin x=-\frac{3}{2}$

$\Rightarrow 3+2(\cos x\cdot \cos y+.....)+2(\sin x\sin y+......)=0$ .........$\left(1\right)$

we can use

$3=1+1+1=({\cos }^{2}x+{\sin }^{2}x)+({\cos }^{2}y+{\sin }^{2}y)+({\cos }^{2}z+{\sin }^{2}z)$

so equation $\left(1\right)$ becomes,

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

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

Now, the sum of two squares can be zero only if each square is itself zero.

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

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