Question

$\cos (x-y)+\cos (y-z)+\cos (z-x)=-\frac{3}{2}$, then $\sum \left(\cos x\right)$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

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

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

$\Rightarrow 2[\cos x\cos y+\sin x\sin y+\cos y\cos z+\sin y\sin z+\cos z.\cos x+\sin z\sin x]+3=0$

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

$\Rightarrow 2[\cos x\cos y+\sin x\sin y+\cos y\cos z+\sin y\sin z+\cos z.\cos x+\sin z.\sin x]+({\sin }^{2}x+{\cos }^{2}x)+({\sin }^{2}y+{\cos }^{2}y)+({\sin }^{2}z+{\cos }^{2}z)=0$

$\Rightarrow {\cos }^{2}x+{\cos }^2+{\cos }^{2}z+2[\cos x.\cos y+\cos y.\cos z+\cos z.\cos x]+{\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z+2[\sin x.\sin y+\sin y.\sin z+\sin z.\sin x]=0$

$\Rightarrow (\cos x+\cos y+\cos z{)}^2+(\sin x+\sin y+\sin z{)}^2=0$

$[\because (a+b+c{)}^2+a^2+b^2+c^2+2(ab+bc+ca)]$

As $\sin x,\sin y,\sin z,\cos x,\cos y\cos z,$ are all real values then sum of two squares will be zero only if both of them are zero

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

$\Rightarrow \sum \cos x=\sum \sin x=0$

$\therefore \sum \left(\cos x\right)=0$