Proving $ \sin{x} - \sin{y} = \sin{(x - y)} \cdot \sqrt{\frac{1 + \cos{(x + y)}} {1 + \cos{(x - y)}}} $

Question

I want to prove following trig identity:

$$ \sin{x} - \sin{y} = \sin{(x - y)} \cdot \sqrt{\frac{1 + \cos{(x + y)}} {1 + \cos{(x - y)}}} $$

for

$$ 0 < x < \pi, 0 < y < \frac{\pi}{2}$$

What would be a strategy to do this? Squaring both sides and using the regular addition/subtraction identities doesn't work. Thanks in advance.

Answer 1

Hint: You will need that $$\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$ $$\sin(x-y)=2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ and $$\cos(x+y)=2\cos^2\left(\frac{x+y}{2}\right)-1$$ $$\cos(x-y)=2\cos^2\left(\frac{x-y}{2}\right)-1$$

Answer 2

$$\dfrac{\sin x-\sin y}{\sin(x-y)}=\dfrac{2\sin(x-y)/2\cos(x+y)/2}{2\sin(x-y)/2\cos(x-y)/2}$$

We have $-\dfrac\pi2<(x\pm y)/2<\dfrac\pi2$

Use $\cos2z=2\cos^2z-1$

$\implies\cos z=+\sqrt{?}$