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 $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}$.

$\frac {d}{da} \sin a = \lim_\limits{x\to 0} \frac {\sin (x+a) - \sin a}{x} = \cos a\\
\lim_\limits{x\to 0} \frac {\sin (x+a) - \sin a}{\sin 2x} = \lim_\limits{x\to 0} \frac {(\sin (x+a) - \sin a)x}{x\sin 2x} = \lim_\limits{x\to 0} (\frac {\sin (x+a) - \sin a}{x})(\frac {x}{\sin 2x})  = \frac 12 \cos a $ 
