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 $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}$.
 A: $\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 $ 
A: Hint: use that $$\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$
A: Hint:
Rewrite the quotient as 
$$\frac{\sin(x+a)-\sin a}x \,\frac{x}{\sin 2x}=\frac{\sin(x+a)-\sin a}x \,\frac{2x}{\sin 2x}\frac 12,$$
and observe the first fraction is but a rate of variation from the value $a$.
A: Write the limit in the following fashion namely
$$
\lim_{x \to 0} \frac{\sin(x+a) - \sin(a)}{\sin2x}=\frac{\lim_{x\to 0}\frac{\sin(a+x)-\sin(a)}{x}}{2\lim_{x\to 0}\frac{\sin 2x}{2x}}=\frac{\cos a}{2}
$$
A: You're doing just fine with your approach. Break it into a sum of limits. The first term is easy. For the second, you need to know
$$\lim_{x\to 0}\frac{\cos x - 1}{x} = \lim_{x\to 0}\frac{(\cos x-1)(\cos x + 1)}{x(\cos x+1)} = - \lim_{x\to 0}\frac{\sin^2 x}{x}\cdot\frac 1{\cos x+1}.$$ Can you finish from this?
A: 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
$$\lim_{x\to0}\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}=\frac{\cos a}{2\times1}-0=\frac{\cos a}{2}$$
A: 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 $h$.
