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

Jun 27, 2016

See below

Explanation:

Consider two vectors representing parallelogram sides

$\vec{a} = \left\{0 , \cos \left(A\right) , \sin \left(A\right)\right\}$
$\vec{b} = \left\{0 , \cos \left(B\right) , \sin \left(B\right)\right\}$

We know that the parallelogram area is given by

$S = \left\lVert \vec{a} \right\rVert \cdot \left\lVert \vec{b} \right\rVert \cos \left(B - A\right) = \left\langle\vec{a} , \vec{b}\right\rangle$

then

$\cos \left(B - A\right) = \frac{\left\langle\vec{a} , \vec{b}\right\rangle}{\left\lVert \vec{a} \right\rVert \cdot \left\lVert \vec{b} \right\rVert}$

substituting we obtain

$\cos \left(B - A\right) = C o s \left(A\right) C o s \left(B\right) + S \in \left(A\right) S \in \left(B\right)$

We can also obtain the parallelogram area using the cross product.

$S = \left\lVert \vec{a} \right\rVert \cdot \left\lVert \vec{b} \right\rVert \sin \left(B - A\right) = \left\langle\hat{i} , \vec{a} \times \vec{b}\right\rangle = C o s \left(A\right) S \in \left(B\right) - C o s \left(B\right) S \in \left(A\right)$

so

$\sin \left(B - A\right) = C o s \left(A\right) S \in \left(B\right) - C o s \left(B\right) S \in \left(A\right)$

Here $\left\lVert \vec{a} \right\rVert = \left\lVert \vec{b} \right\rVert = 1$

Note.
$\left\langle\vec{a} , \vec{b}\right\rangle = \left\langle\left\{0 , \cos \left(A\right) , \sin \left(A\right)\right\} , \left\{0 , \cos \left(B\right) , \sin \left(B\right)\right\}\right\rangle = C o s \left(A\right) C o s \left(B\right) + S \in \left(A\right) S \in \left(B\right)$
$\vec{a} \times \vec{b} = \left\{0 , \cos \left(A\right) , \sin \left(A\right)\right\} \times \left\{0 , \cos \left(B\right) , \sin \left(B\right)\right\} = \left\{C o s \left(A\right) S \in \left(B\right) - C o s \left(B\right) S \in \left(A\right) , 0 , 0\right\}$