Question

If $0<x,y<\pi$ and $\cos x+\cos y-\cos (x+y)=\frac{3}{2},$ then $\sin x+\cos y$ is equal to :

• A. $\frac{1+\sqrt{3}}{2}$
• B. $\frac{1-\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1+\sqrt{3}}{2}$

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$
$\Rightarrow 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-[2{\cos }^2\left(\frac{x+y}{2}\right)-1]=\frac{3}{2}$
$\Rightarrow 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-2{\cos }^2\left(\frac{x+y}{2}\right)=\frac{1}{2}$
$\Rightarrow 4\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-4{\cos }^2\left(\frac{x+y}{2}\right)=1={\cos }^2\frac{(x-y)}{2}+{\sin }^2\left(\frac{x-y}{2}\right)$
$\Rightarrow 4{\cos }^2\left(\frac{x+y}{2}\right){\cos }^2\left(\frac{x-y}{2}\right)-4\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)+{\sin }^2\left(\frac{x-y}{2}\right)=0$
$\Rightarrow {(\cos \left(\frac{x-y}{2}\right)-2\cos \left(\frac{x+y}{2}\right))}^2+{\sin }^2\left(\frac{x-y}{2}\right)=0$
$\Rightarrow \sin \frac{x-y}{2}=0\Rightarrow x=y$
and $\cos \frac{x-y}{2}=2\cos \frac{x+y}{2}$
$\Rightarrow \cos x=\frac{1}{2}=\cos y$
$\therefore \sin x+\cos y=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}$