Help me prove: sin(A+B) = sinA cosB + cosA sinB Can you help me prove that: sin(A+B) = sinA cosB + cosA sinB?
Thanks! 
 A: Hint: write it as $\sin(x) = 1/2 i e^{-i x} - 1/2 i e^{i x}$
A: I found a very interesting geometric proof here.
It proves it relatively clearly and concisely.
A: Here are  two for the price of one (using Euler's formula):
$$\color{red}{\cos(A+B)}+i\color{green}{\sin(A+B)} \equiv e^{i(A+B)} \equiv e^{iA} \times e^{iB}$$
$$\equiv [\cos(A)+i\sin(A)][\cos(B)+i\sin(B)]$$
$$\equiv \color{red}{[\cos(A)\cos(B)-\sin(A)\sin(B)]}+i\color{green}{[\sin(A)\cos(B)+\cos(A)\sin(B)]}$$
Now equate imaginary parts to give the result for $\sin(A+B)$ (and, if you want, equate real parts to give the result for $\cos(A+B)$).
A: Here is a proof without using complex numbers, from Apostol.
$$\sin(x+y) = -\cos(x + y + \frac{\pi}{2}) = -\cos x\cos(y + \frac{\pi}{2}) + \sin(x)\ \sin(y + \frac{\pi}{2})$$
$$= \cos x\ \sin y + \sin x\cos y $$
Assuming you can prove the formula for the cosine, of course, which Apostol gives as a property of the cosine.
A: May be a little unconventional, but assuming we have an isosceles right angled triangle.
Then, $sin 90^\circ=1$
$$=\frac{1}{2}+\frac{1}{2}$$
$$=\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}$$
$$=cos 45^\circ \cdot sin 45^\circ+sin 45^\circ \cdot cos 45^\circ$$
The similar can be proved for a scalene triangle as well.
