Question

If $\cos x\cos y-\cos (x+y)=\frac{3}{2}$, then

• A. $x+y=0$
• B. $x=2y$
• C. $x=y$
• D. $2x=y$

Answer 1

Answer: (C)

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$
or $2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-2{\cos }^2\left(\frac{x+y}{2}\right)+1=\frac{3}{2}$
or $2{\cos }^2\left(\frac{x+y}{2}\right)-2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)+\frac{1}{2}=0$
Now, $\cos \left(\frac{x+y}{2}\right)$ is always real, then discriminant $\ge 0$ .
Thus, $4{\cos }^2\left(\frac{x-y}{2}\right)-4\ge 0$
or ${\cos }^2\left(\frac{x-y}{2}\right)\ge 1$
or ${\cos }^2\left(\frac{x-y}{2}\right)=1$
or $\frac{x-y}{2}=0$ or $x=y$