Proof of trigonometric identity using vector calculus

Question

Question: Using vector calculus, show that $\sin (A+B) = \sin A \cos B + \cos A \sin B$

I have no idea how to even attempt the question. A small hint to help me get started would be greatly appreciated!

Answer 1

Consider the unit vectors $\langle \cos(90-A), \sin(90-A)\rangle $ and $\langle \cos B, \sin B \rangle $

Take the dot product $$\langle \cos(90-A), \sin(90-A)\rangle \bullet \langle \cos B, \sin B \rangle = \cos (90 -A -B) $$

Answer 2

Hint:

The matrix $\begin{pmatrix}\cos B&-\sin B\\\sin B&\cos B\end{pmatrix}$ is a counter-clockwise rotation by angle $B$. Apply it (via left-multiplication) to the point $\begin{pmatrix}\cos A\\\sin A\end{pmatrix}$. The result is $\begin{pmatrix}\cos (A+B)\\\sin (A+B)\end{pmatrix}$. But $$ \begin{pmatrix}\cos B&-\sin B\\\sin B&\cos B\end{pmatrix}\begin{pmatrix}\cos A\\\sin A\end{pmatrix}=\begin{pmatrix}\cos A\cos B-\sin A\sin B\\\sin A\cos B+\cos A\sin B\end{pmatrix}. $$