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

1 Answer
Jun 27, 2016

Answer:

See below

Explanation:

Consider two vectors representing parallelogram sides

#vec a = {0,cos(A),sin(A)}#
#vec 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 >>#

then

#cos(B-A) = << vec a, vec b >>/(norm (vec a)cdot norm (vec b))#

substituting we obtain

#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)#

so

#sin(B-A)=Cos(A) Sin(B)-Cos(B) Sin(A)#

Here #norm (vec a)=norm(vec 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)#
#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}#