Consider two vectors representing parallelogram sides
vec a = {0,cos(A),sin(A)}→a={0,cos(A),sin(A)}
vec b = {0,cos(B),sin(B)}→b={0,cos(B),sin(B)}
We know that the parallelogram area is given by
S = norm (vec a)cdot norm (vec b) cos(B-A) = << vec a, vec b >>S=∥∥→a∥∥⋅∥∥∥→b∥∥∥cos(B−A)=⟨→a,→b⟩
then
cos(B-A) = << vec a, vec b >>/(norm (vec a)cdot norm (vec b))cos(B−A)=⟨→a,→b⟩∥∥→a∥∥⋅∥∥∥→b∥∥∥
substituting we obtain
cos(B-A) =Cos(A) Cos(B) + Sin(A) Sin(B)cos(B−A)=cos(A)cos(B)+sin(A)sin(B)
We can also obtain the parallelogram area using the cross product.
S = norm(vec a)cdot norm(vec b) sin(B-A) =<< hat i ,veca xx vec b>> = Cos(A) Sin(B)-Cos(B) Sin(A)S=∥∥→a∥∥⋅∥∥∥→b∥∥∥sin(B−A)=⟨ˆi,→a×→b⟩=cos(A)sin(B)−cos(B)sin(A)
so
sin(B-A)=Cos(A) Sin(B)-Cos(B) Sin(A)sin(B−A)=cos(A)sin(B)−cos(B)sin(A)
Here norm (vec a)=norm(vec b) = 1∥∥→a∥∥=∥∥∥→b∥∥∥=1
Note.
<< vec a, vec b >> =<< {0,cos(A),sin(A)}, {0,cos(B),sin(B)} >> = Cos(A) Cos(B) + Sin(A) Sin(B)⟨→a,→b⟩=⟨{0,cos(A),sin(A)},{0,cos(B),sin(B)}⟩=cos(A)cos(B)+sin(A)sin(B)
vec a xx vec b = {0,cos(A),sin(A)} xx {0,cos(B),sin(B)} = {Cos(A) Sin(B)-Cos(B) Sin(A),0,0}→a×→b={0,cos(A),sin(A)}×{0,cos(B),sin(B)}={cos(A)sin(B)−cos(B)sin(A),0,0}
