Question

If $\sin x+\sin y+\sin z=0=\cos x+\cos y+\cos z,$ then find the value of expression $\cos (y-x)+\cos (z-y)+\cos (x-z)$.

Answer 1

Given, $\sin x+\sin y+\sin z=0$ and $\cos x+\cos y+\cos z=0$

$⟹$ $\sin x+\sin y=-\sin z$ and $\cos x+\cos y=-\cos z$

${\left(\sin x+\sin y\right)}^2={\sin }^{2}z$ and ${\left(\cos x+\cos y\right)}^2={\cos }^{2}z$

Now, ${\left(\sin x+\sin y\right)}^2+{\left(\cos x+\cos y\right)}^2={\sin }^{2}z+{\cos }^{2}z$

$\Rightarrow {\sin }^{2}x+{\sin }^{2}y+2\sin x\sin y+{\cos }^{2}x+{\cos }^{2}y+2\cos x\cos y=1$ ...... $[sin²\theta +cos²\theta =1]$

$\Rightarrow 1+1+2(\sin x\sin y+\cos x\cos y)$

$\Rightarrow 1+1+2\cos (x-y)=1$ .................$\cos (a-b)=\cos a\cdot \cos b+\sin a\cdot sinb$

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

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

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