How do you derive the formula of $\sin (a+b)$ from $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ I know that cos is even while sin is odd, and I know $\cos(\pi)=\sin((\pi/2)-x)$, but I  still can't figure the derivation of $\sin (a+b)$ from $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. Could you give me any hint?
 A: $$\sin(a+b)=\cos\left(\left(\frac{\pi}{2}-a\right)+(-b)\right)=\cos\left(\frac{\pi}{2}-a\right)\cos(-b)-\sin\left(\frac{\pi}{2}-a\right)\sin(-b)\\=\sin(a)\cos(-b)-\cos(a)(-\sin(b))=\sin(a)\cos(b)+\cos(a)\sin(b)$$
A: A very short hint:
Deriving ;o)
and some details:
Start from
$$\cos(a+b)=\cos a \cos b-\sin a\sin b$$
and differentiate w. r. t., say, $a$:
$$-\sin(a+b)=-\sin a\cos b-\cos a \sin b.$$
A: Since Element118 has already given what you expected, here is a geometric proof (the grey squares represent right angle). 

By definition, we know that $$\sin(\alpha+\beta)=\frac{|BC|+|CD|}{|AB|}=\frac{|BC|}{|AB|}+\frac{|CD|}{|AB|}$$ 
We want $$\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha)$$ 
We have $\cos(\alpha)=\frac{|BC|}{|BE|}$ and $\sin(\beta)=\frac{|BE|}{|AB|}$ 
and it gives:
\begin{equation*}
\frac{|BC|}{|AB|}=\frac{|BC|\cdot|BE|}{|AB|\cdot|BE|}=\underbrace{\frac{|BC|}{|BE|}}_{\cos(\alpha)}\underbrace{\frac{|BE|}{|AB|}}_{\sin(\beta)}
\end{equation*}
We repeat this for $\frac{|CD|}{|AB|}$ and we get
\begin{equation}
\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha)
\end{equation}
