Question

If $0<x,y<π$ 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)

$\cos x+\cos y−\cos (x+y)=\frac{3}{2}$
${\cos }^2\left(\frac{x+y}{2}\right)−\cos \left(\frac{x+y}{2}\right)â‹\dots \cos \left(\frac{x−y}{2}\right)$
$+\frac{1}{4}â‹\dots {\cos }^2\left(\frac{x−y}{2}\right)+\frac{1}{4}{\sin }^2\left(\frac{x−y}{2}\right)=0$
$⇒{\left(\cos \left(\frac{x+y}{2}\right)−\frac{1}{2}\cos \left(\frac{x−y}{2}\right)\right)}^2+\frac{1}{4}{\sin }^2\left(\frac{x−y}{2}\right)=0$
$⇒\sin \left(\frac{x−y}{2}\right)=0$ and
$\cos \left(\frac{x+y}{2}\right)=\frac{1}{2}\cos \left(\frac{x−y}{2}\right)$
$⇒x=y$ and $\cos x=\frac{1}{2}=\cos y$
$∴\sin x=\frac{\sqrt{3}}{2}$
$⇒\sin x+\cos y=\frac{1+\sqrt{3}}{2}$