Show that a.b = (1/4) |a+b|^2 - (1/4) |a-b|^2

Let us take the two vectors a and
b.

Their dot product is |a|*|b|*cos D, where D is the angle
between the vectors.

Let the vector a make an angle of A
with the horizontal and vector b make an angle of B with the
horizontal.

(1/4)*|a + b|^2 - (1/4)*|a -
b|^2

=> (1/4)[|a + b|^2 - (1/4)*|a -
b|^2]

=> (1/4)[[sqrt((|a|*sin A + |b|*sin B)^2 +
(|a|*cos A + |b|*cos B)^2)]^2 - [sqrt((|a|*sin A - |b|*sin B)^2 + (|a|*cos A - |b|*cos
B)^2)]^2 ]]

=> (1/4)[[sqrt(|a|^2*(sin A)^2 +
|b|^2*(sin B)^2 + 2|a|*|b|*sin A * sin B+ |a|^2*(cos A)^2 + |b|^2*(cos B)^2 +
2|a|*|b|*cos A * cos B]^2 -  [sqrt((|a|^2*(sin A)^2 + |b|^2*(sin B)^2 - 2|a|*|b|*sin A *
sin B+ |a|^2*(cos A)^2 + |b|^2*(cos B)^2 - 2|a|*|b|*cos A *
cosB]^2]

simplify using (cos x)^2 + (sin x)^2 =
1

=> (1/4)[[(|a|^2 + 2|a|*|b|*sin A * sin B +
2|a|*|b|*cos A * cos B] -  [(|a|^2 + |b|^2 - 2|a|*|b|*sin A * sin B - 2|a|*|b|*cos A *
cos B]]

=> (1/4)[4|a|*|b|*sin A * sin B +
4|a|*|b|*cos A * cos B]

=> [|a|*|b|*sin A * sin B +
|a|*|b|*cos A * cos B]

use cos (A - B) = cos A * cos B +
sin A * sin B

=> |a|*|b|*cos ( A -
B)

=> |a|*|b|*cos
D

which is
a.b

Therefore we prove that a.b = (1/4)*|a +
b|^2 - (1/4)*|a - b|^2