Question

If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z,$ then

• A. $\cos \frac{x-y}{2}=\pm \frac{\sqrt{3}}{2}$
• B. $\cos \frac{x-y}{2}=\pm \frac{1}{2}$
• C. ${\sin }^2\frac{x-y}{2}=\frac{3}{4}$
• D. $\tan \frac{x-y}{2}=\pm \frac{1}{\sqrt{3}}$

Answer 1

Answer: (B), (C)

The correct options are
B $\cos \frac{x-y}{2}=\pm \frac{1}{2}$
C ${\sin }^2\frac{x-y}{2}=\frac{3}{4}$

$\cos x+\cos y+\cos z=0$
and $\sin x+\sin y+\sin z=0$
$\Rightarrow \cos x+\cos y=-\cos z$
and $\sin x+\sin y=-\sin z$
$\Rightarrow 2\cos \frac{x+y}{2}\cos \frac{x-y}{2}=-\cos z\cdots \left(1\right)$
and $2\sin \frac{x+y}{2}\cos \frac{x-y}{2}=-\sin z\cdots \left(2\right)$

$(1{)}^2+(2{)}^2$ gives,
$4{\cos }^2\frac{x-y}{2}{\cos }^2\frac{x+y}{2}+4{\cos }^2\frac{x-y}{2}{\sin }^2\frac{x+y}{2}={\cos }^{2}z+{\sin }^{2}z$
$\Rightarrow 4{\cos }^2\frac{x-y}{2}[{\cos }^2\frac{x+y}{2}+{\sin }^2\frac{x+y}{2}]={\cos }^{2}z+{\sin }^{2}z$
$\Rightarrow 4{\cos }^2\frac{x-y}{2}=1$
$\therefore {\cos }^2\frac{x-y}{2}=\frac{1}{4}$
$\Rightarrow \cos \frac{x-y}{2}=\pm \frac{1}{2}$

Since ${\cos }^2\frac{x-y}{2}=\frac{1}{4},$
$1-{\sin }^2\frac{x-y}{2}=\frac{1}{4}$
$\Rightarrow {\sin }^2\frac{x-y}{2}=\frac{3}{4}$