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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!

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I drew two vectors that make an angle A and B with their resultant. I tried taking components of one of the vectors along the other, but that didn't help me :/ @pbs – Gummy bears Sep 1 '14 at 15:42
The vector calculus approach can be seen as a translation of $\cos(A+B)+i\sin(A+B)=e^{i(A+B)}=e^{iA}e^{iB}=(\cos(A)+i\sin(A))(\cos(B)+i\sin(B‌​))$. – Kim Jong Un Sep 1 '14 at 15:44
How are you using vectors to represent sine and consine? Also you are missing a $ – graydad Sep 1 '14 at 15:44
@user166967 Ummm I drew two vectors that make an angle with their resultant? So taking components you get sine and cosine. And thanks for the heads up – Gummy bears Sep 1 '14 at 15:46
@KimJongUn Ummm vector analysis would be a better term I guess? – Gummy bears Sep 1 '14 at 15:46

2 Answers 2

up vote 3 down vote accepted

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

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

enter image description here

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Hmmmm..... So shouldn't I take the cross product? As I need sine? – Gummy bears Sep 1 '14 at 16:03
That works too ! But the dot product is a bit simple to visualize for me :) – ganeshie8 Sep 1 '14 at 16:05
$$(a_1,a_2) \bullet (b_1, b_2) = a_1b_1 + a_2b_2$$ – ganeshie8 Sep 1 '14 at 16:06
$\cos(90-A-B) = \cos (90 - (A+B)) = \sin(A+B) $ – ganeshie8 Sep 1 '14 at 16:08
Indeed it does use polar form and the euler's definition. its a simple proof : $$ \begin{align} \\ \sin(A+B) &= \mathcal{Img} \left(e^{i (A+B)}\right)\\&= \mathcal{Img} \left(e^{iA} . e^{iB}\right) \\&= \mathcal{Img} \left( (\cos A + i\sin A) . (\cos B + i\sin B)\right) \\&= \mathcal{Img} \left( \cos A \cos B - \sin A \sin B + i(\sin A \cos B + \cos A \sin B)\right) \\&= \sin A \cos B + \cos A \sin B \end{align} $$ – ganeshie8 Sep 1 '14 at 16:25


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 $$ \begin{pmatrix}\cos B&-\sin B\\\sin B&\cos B\end{pmatrix}\begin{pmatrix}\cos A\\\sin A\end{pmatrix}=\begin{pmatrix}\cos A\cos B-\sin A\sin B\\\sin A\cos B+\cos A\sin B\end{pmatrix}. $$

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Don't understand it I'm afraid. – Gummy bears Sep 1 '14 at 15:51
I expanded the post slightly. For intuition, draw the point $(\cos A,\sin A)$ in the plane. What will happen after you rotate that point counterclockwise by the angle $B$? – Kim Jong Un Sep 1 '14 at 15:54
Ohhh.... I see what you did there now. But where did you get the matrix from? Any proof for that/ – Gummy bears Sep 1 '14 at 16:00
See – Kim Jong Un Sep 1 '14 at 16:04
I see now. Very ingenious way of doing it. Any easier way to do it though? – Gummy bears Sep 1 '14 at 16:05

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