Question

$\cos x+\cos y+cosz$ $=0=$$\sin x+\sin y+\sin z$. Find the value of ${\cos }^2\left(\frac{x-y}{2}\right)=$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

Answer: (B)

$\frac{1}{4}$

$\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$
$\Rightarrow \cos x+\cos y=-\cos z$ ...(1)
$\sin x+\sin y=-\sin z$ ...(2)
Squaring and adding eq. $\left(1\right)$ & eq. $\left(2\right)$ we get,
${\cos }^{2}x+{\cos }^{2}y+2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y+2\sin x\sin y$
$={\cos }^{2}z+{\sin }^{2}z\Rightarrow 1+1+2(\cos x\cos y+\sin x\sin y)=1$
$\Rightarrow \cos (x-y)=-\frac{1}{2}$
$\Rightarrow 2{\cos }^2\left(\frac{x-y}{2}\right)-1=-\frac{1}{2}$
$\Rightarrow {\cos }^2\left(\frac{x-y}{2}\right)=\frac{1}{4}$