Question

For the smallest positive values of 'x' and 'y' the equation $2(sinx+siny)-2cos(x-y)=3$ has a solution then which of the following is/are true

• A. $sin\left(\frac{x+y}{2}\right)=1$
• B. $cos\left(\frac{x-y}{2}\right)=\frac{1}{2}$
• C. Number of ordered pairs (x, y) is 2
• D. Number of ordered pairs (x, y) is 3

Answer 1

The correct option is A $sin\left(\frac{x+y}{2}\right)=1$

$2(sinx+siny)-2cos(x-y)=3$
$sinC+sinD=2sin\frac{C+D}{2}cos\frac{C-D}{2}$
$4sin\frac{x+y}{2}cos\frac{x-y}{2}-2[2co{s}^2\frac{x-y}{2}-1]=3$
$4sin\frac{x+y}{2}cos\frac{x-y}{2}-4co{s}^2\frac{x-y}{2}+2=3$
$4cos\frac{(x-y)}{2}[sin\frac{x+y}{2}-cos\frac{x-y}{2}]=3-2=1$
$4cos\frac{x-y}{2}[sin\frac{x+y}{2}-cos\frac{x-y}{2}]=1$