Question

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

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

Hint:

When a trigonometric equation seems complex, try to rearrange it into a quadratic form for one of the trigonometric functions. Then, use the condition that the discriminant must be non-negative for real solutions. This can often constrain the variables and lead to a solution.

Answer 1

The Correct Option is B

Given: \[ \cos x + \cos y - \cos(x+y) = \frac{3}{2} \] Using $\cos(x+y)=\cos x\cos y-\sin x\sin y$, \[ 2\cos x - (2\cos^2 x - 1)=\frac{3}{2} \] \[ 4\cos^2 x - 4\cos x + 1=0 \] \[ (2\cos x - 1)^2=0 \Rightarrow \cos x=\frac{1}{2} \Rightarrow x=y=\frac{\pi}{3} \] \[ \sin x=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \] \[ \boxed{\frac{\sqrt{3}}{2}} \]