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

Answer 1

Answer: (B)

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