Confusion regarding the derivation of $\cos(x)$ when differentiating $\sin(x)$ The textbook I'm reading derives it like this:
$\eqalign{
  & y = \sin x \Rightarrow \left( 1 \right)  \cr 
  & y + \delta y = \sin (x + \delta x) \Rightarrow (2) \cr} $
Subtracting equation (1) from (2):
$\eqalign{
  & y + \delta y - y = \sin (x + \delta x) - \sin x  \cr 
  & \delta y = \sin (x + \delta x) - \sin x  \cr 
  &  = 2\cos (x + {1 \over 2}\delta x)\sin \left({1 \over 2}\delta x\right) \cr} $

Without completing the whole "proof" I can tell you where I am confused.
$\delta y = \sin (x + \delta x) - \sin x$
to this:
$2\cos (x + {1 \over 2}\delta x)\sin \left({1 \over 2}\delta x\right)$ 
What happened here? It looks like the double angle identity was used?
Isn't the double angle formula used for when two angles are the same? I know dx is very small, but still, the value of x has been altered so shouldn't the "angle addition" identity be used instead? Further more what happened to the -sin(x)? How was it incorporated into $2\cos (x + {1 \over 2}\delta x)\sin \left({1 \over 2}\delta x\right)$ ?
This is confusing me quite a bit, any help would be greatly appreciated, thank you.
EDIT: Furthermore, 1/2(dx): where does this come from?
 A: There was used a well-known formula $$\sin{\alpha}-\sin{\beta}=2\cos{\dfrac{\alpha+\beta}{2}}\sin{\dfrac{\alpha-\beta}{2}}.$$
A: To prove that formula just use
Let $\alpha = A+B, \beta = A-B$. Then note that $\alpha +\beta =2A$ and $\alpha - \beta=2B$. Now you have $A=\displaystyle \frac{\alpha+\beta}{2}$ and $B=\displaystyle \frac{\alpha-\beta}{2}$
\begin{align*}
\sin(\alpha)-\sin(\beta) &= \sin(A+B) - \sin(A-B) 
\\ &= \sin(A)\cdot\cos(B) + \cos(A)\cdot \sin(B) - \sin(A)\cdot\cos(B)+\cos(A)\cdot\sin(B) \\ &=2 \cdot \cos(A) \cdot \sin(B) \\
&= 2 \cdot \cos\left(\frac{\alpha+\beta}{2}\right) \cdot \sin\left(\frac{\alpha-\beta}{2}\right) 
\end{align*}

$\Large\text{Method 2}$
\begin{align*}
f'(x) &= \lim_{\delta x\to 0} \frac{\sin(x+\delta{x}) -\sin(x)}{\delta{x}} \\ &=\lim_{\delta{x}\to0} \frac{\sin(x)\cdot\cos(\delta{x}) +\cos(x)\cdot\sin\delta{x} -\sin(x)}{\delta{x}} \\ &=\lim_{\delta{x}\to0} \frac{\sin(x)\cdot \bigl(\cos(\delta{x})-1\bigr)}{\delta{x}} + \lim_{\delta{x}\to 0}\frac{\cos{x}\cdot \sin(\delta{x})}{\delta{x}}
\\ &= \lim_{\delta{x}\to 0} \frac{\sin(x)\cdot \bigl(\cos(\delta{x})-1\bigr)}{\delta{x}} + \cos(x) \qquad \Bigl[\because \small{\lim_{x\to 0} \frac{\sin{x}}{x}=1} \Bigr]
\end{align*}
Now i leave it to you to prove $\displaystyle\lim_{\delta{x}\to 0}\frac{\sin(x)\cdot \bigl(\cos(\delta{x})-1\bigr)}{\delta{x}} =0$. This can be proved by observing that $1-\cos(2x) = 2\cdot \sin^{2}(x)$.
A: Recall the identities $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and its close relative $\sin(A-B)=\sin A\cos B-\cos A\sin B$. Subtract. We get
$$\sin(A+B)-\sin(A-B)=2\cos A\sin B.$$
Let $A+B=s$ and $A-B=t$. Then $A=\frac{s+t}{2}$ and $B=\frac{s-t}{2}$. So we obtain
$$\sin s-\sin t=2\cos\left(\frac{s+t}{2} \right) \sin\left(\frac{s-t}{2}  \right).$$
Remark: We give a somewhat different derivation of the derivative formula, that I think is a little easier for students. We also change notation slightly for ease of typing. 
We are interested in the behaviour of $\frac{\sin(x+h)-\sin x}{h}$ as $h$ approaches $0$. The top is $\sin x\cos h=\cos x \sin h$. Divide by $h$ and rearrange slightly. We get
$$\cos x\frac{\sin h}{h} -\sin x\frac{1-\cos h}{h}.$$
Let $h\to 0$. What happens to the first part is familiar. As for $\frac{1-\cos h}{h}$, multiply top and bottom by $1+\cos h$. We get
$$\frac{1}{1+\cos h}\frac{\sin^2 h}{h}.$$
Since $\frac{\sin h}{h}$ has limit $1$, we conclude that $\frac{\sin^2 h}{h}$ has limit $0$. 
