How do I derive the identity $\sin(a+b)=\sin(a)\cos(b) + \cos(a)\sin(b)$ using the Unit Circle I'd like it explained through the unit circle as I find trig identities easier much easier to understand in this manner. 
EDIT: I know you have to apply the identity $\sin(x)=\cos(90-x)$, but I'm wondering how i'd visualise all this on the unit circle?
 A: Using complex numbers and exponential form perhaps help (at least algebraically) to digest these trigonometric addition formulas:
All we have to know is $\cos a+i\cdot\sin a=e^{ai}$ for any $a\in\Bbb R$, and that $i^2=-1$, and that $e^{x+y}=e^x\cdot e^y$ for any $x,y\in\Bbb C$.
Then calculate both sides of $e^{(a+b)i}=e^{ai}e^{bi}$.
If you prefer, instead, you can use the matrices of rotation:
$$R_a:=\pmatrix{\cos a&-\sin a\\ \sin a &\cos a}$$
and use matrix multiplication to verify the identities, knowing that
$$R_{a+b}=R_a\cdot R_b \ .$$
A: This video will clear matters beautifully.
Make sure to watch it, and then the next.
A: The explanation given by eepsmedia here is what you are looking for. Although his argument gives the cosine addition formula, and only in the case when $\alpha+\beta < \pi/2$, you should be able to use the same methods to obtain the sine angle addition formula.
A: If the reader knows a little calculation, we have an elegant proof of this identity. Let 
$$
f(x) = \sin(x + b) - \sin x\cos b - \sin b\cos x \ \Rightarrow 
$$
$$
f^{\prime}(x) = \cos(x + b) - \cos x \cos b + \sin b \sin x = \cos(x + b) - \cos(x + b) = 0
$$
since $\cos(x + b) = \cos x \cos b - \sin b \sin x$. Thus, 
$$
f(x) = \sin(x + b) - \sin x\cos b - \sin b\cos x = C, \quad C \in \mathbb{R}
$$
For $x = 0$, we have $C = 0$. In particular, at $x = a$,
$$
\boxed{\sin(a + b) = \sin a\cos b + \sin b\cos a}
$$
