Question

The smallest positive values of x and y that satisfy 2(sinx + sin y) - 2 cos (x - y) = 3 is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given equation is,

or $2.2sin\left(\frac{x+y}{2}\right)cos\left(\frac{x-y}{2}\right)-2[2co{s}^2\left(\frac{x-y}{2}\right)-1]=3$

or $4sin\left(\frac{x+y}{2}\right).cos\left(\frac{x-y}{2}\right)-4co{s}^2\left(\frac{x-y}{2}\right)=1$

or $4cos\left(\frac{x-y}{2}\right).[sin\left(\frac{x+y}{2}\right)-cos\left(\frac{x-y}{2}\right)]=1$

or $4cos\left(\frac{x-y}{2}\right).[sin\left(\frac{x+y}{2}\right)-sin(\frac{\pi }{2}-\frac{x-y}{2})]=1$

or $4cos\left(\frac{x-y}{2}\right).2cos\left(\frac{\pi +2y}{4}\right).sin\left(\frac{2x-\pi }{4}\right)=1$

or $cos\left(\frac{x-y}{2}\right)cos(\frac{\pi }{4}+\frac{y}{2})sin(\frac{x}{2}-\frac{\pi }{4})=\frac{1}{8}={\left(\frac{1}{2}\right)}^3$

2 (sin x + sin y) - 2cos (x - y) = 3

If x and y be positive and smallest, then

$cos\left(\frac{x-y}{2}\right)=cos(\frac{\pi }{4}+\frac{y}{2})=sin(\frac{x}{2}-\frac{\pi }{4})=\frac{1}{2}$

$\theta \frac{\pi }{4}+\frac{y}{2}=\frac{\pi }{3},\frac{x}{2}-\frac{\pi }{4}=\frac{\pi }{6}$

$\Rightarrow \frac{y}{2}=\frac{\pi }{12},\frac{x}{2}=\frac{5\pi }{12},\therefore y=\frac{\pi }{6},x=\frac{5\pi }{6}$

Which satisfy $cos\left(\frac{x-y}{2}\right)=\frac{1}{2}$

Hence required solution is

$x=\frac{5\pi }{6}$ and y = $\frac{\pi }{6}$.