Evaluation of $\lim_{x \to 0} \frac{\sin(x+a) -\sin(a)}{\sin2x}$. I am having trouble proving out the below:

$$\lim_{x \to 0} \frac{\sin(x+a) - \sin(a)}{\sin2x}$$

I have got as far as below using $\sin(a) + \sin(b)$ in numerator and $\sin(2a)$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$$\lim_{x \to 0}\frac{\sin(x)\cos(a) + \sin(a)[\cos(x) - 1]}{2\sin(x)\cos(x)}$$
I need to simplify it down to apply $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}$.

Continue with this:
$$\frac{\sin x \cos a}{2\sin x\cos x} + \frac{\sin a[\cos x - 1]}{2\sin x\cos x}$$
$$\frac{\cos a}{2\cos x} - \sin a\frac{2\sin^2\frac{x}{2}}{4\sin\frac{x}{2}\cos\frac{x}{2}\cos x}$$
$$\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}$$
then
$$\lim_{x\to0}\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}=\frac{\cos a}{2\times1}-0=\frac{\cos a}{2}$$
