Help me prove: sin(A+B) = sinA cosB + cosA sinB Can you help me prove that: sin(A+B) = sinA cosB + cosA sinB?

Thanks!
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 $\cos(A+B)$).
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.