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

Use the sum to product formula 
$$\sin(x)-\sin(y) = 2\cdot \cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$\begin{align}
\sin(x+a)-\sin(a)~&=~2\cdot \cos\left(\frac{(x+a)+a}2\right)\sin\left(\frac{(x+a)-a}2\right)\\
&=~2\cdot\cos\left(\frac{x}2+a\right)\sin\left(\frac{x}2\right)
\end{align}$$
Subsitute this into the limit yields to
$$\begin{align}
\lim_{x\to 0}\frac{2\cdot\cos\left(\frac{x}2+a\right)\sin\left(\frac{x}2\right)}{2\cdot\sin(x)\cos(x)}~&=~\lim_{x\to 0}\frac{\cos\left(\frac{x}2+a\right)\sin\left(\frac{x}2\right)}{\sin(x)\cos(x)}\\
&=~\lim_{x\to 0}\frac{\cos\left(\frac{x}2+a\right)}{\cos(x)}\cdot\lim_{x\to 0}\frac{\sin\left(\frac{x}2\right)}{x}\\
&=~\cos(a)~\cdot\lim_{h\to 0}\frac{\sin\left(h\right)}{2h}\\
&=~\cos(a)~\cdot\frac12\lim_{h\to 0}\frac{\sin\left(h\right)}{h}\\
&=~\frac{\cos(a)}{2}
\end{align}$$
Where in the end $\frac{x}2$ was substituted by $h$.
