Question

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

• A. $\cos x+\cos y-\cos z=0$
• B. $\sin x+\sin y+\sin z=0$
• C. $\sin x+\sin y+\sin z=\cos x+\cos y+\cos z$
• D. $\sin x+\sin y+\sin z=\cos x+\cos y+\cos z=0$

Answer 1

Answer: (B), (C), (D)

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

From the given equation
$2\sum \cos x\cos y+2\sum \sin x\sin y+3=0$
$2\sum \cos x\cos y+2\sum \sin x\sin y+1+1+1=0$
$2\sum \cos x\cos y+2\sum \sin x\sin y+{\sin }^{2}x+{\cos }^{2}x+{\sin }^{2}y+{\cos }^{2}y+{\sin }^{2}z+{\cos }^{2}z=0$
$[{\cos }^{2}x+{\cos }^{2}y+{\cos }^{2}z+2\sum \cos x\cos y]+[{\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z+2\sum \sin x\sin y]=0$
${(\cos x+\cos y+\cos z)}^2+{(\sin x+\sin y+\sin z)}^2=0$
$\Rightarrow \cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$
Hence, option 'D' is correct.