# How will you prove the formula sin(A+B)=sinAcosB+cosAsinB using formula of scalar product of two vectors?

Oct 2, 2016

See below.

#### Explanation:

Given

${\vec{v}}_{\alpha} = \left(\sin \alpha , \cos \alpha\right)$ and
${\vec{v}}_{\beta} = \left(\sin \beta , \cos \beta\right)$

both unit vectors, because

$\left\lVert {\vec{v}}_{\alpha} \right\rVert = \left\lVert {\vec{v}}_{\beta} \right\rVert = 1$

Have their inner product given by

$\left\langle{\vec{v}}_{\alpha} , {\vec{v}}_{\beta}\right\rangle = \sin \alpha \sin \beta + \cos \alpha \cos \beta$ and which is the projection on ${\vec{v}}_{\alpha}$ onto ${\vec{v}}_{\beta}$ or vice-versa.

This is exactly $\cos \left(\alpha - \beta\right) = \cos \left(\beta - \alpha\right)$

If we take complex numbers and using the de Moivre's identity which reads

${e}^{i x} = \cos x + i \sin x$ we have

$\left(\cos \alpha + i \sin \alpha\right) \left(\cos \beta + i \sin \beta\right) = {e}^{i \alpha} {e}^{i \beta}$ $= \cos \alpha \cos \beta - \sin \alpha \sin \beta + i \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) = {e}^{i \left(\alpha + \beta\right)} = \cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)$.

In the case of vectors we have projections and in the case of complex numbers we have rotations.

$\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin \left(\alpha + \beta\right)$