Trigonometric identity problem

Question

$$\frac{\sin(2x)}{2 \sin (x)}-\frac{\cos(2x)}{\cos(x)+\sin(x)}=\sin(x).$$ I got it on a test and want an answer. I always hit a dead end with the identities I learned.

Answer 1

$\sin(2x) = 2\sin(x)\cos(x)$, and $\cos(2x) = \cos^{2}(x) - \sin^{2}(x)$, which factors as $(\cos(x) - \sin(x))(\cos(x) + \sin(x))$. That turns your identity into $$ \frac{2\sin(x)\cos(x)}{2\sin(x)} - \frac{\cos^{2}(x) - \sin^{2}(x)}{\cos(x) + \sin(x)} = \cos(x) - (\cos(x) - \sin(x)) = \sin(x).$$

Answer 2

Hints:

1. Use $\sin(2x) = 2 \sin(x) \cos(x).$
2. Use $\cos(2x) = \cos^2 (x) - \sin^2(x).$
3. Factor $\cos^2 (x) - \sin^2(x) = (\cos(x) + \sin(x))(\cos (x) - \sin(x)).$