prove that:sinAcosB+cosAsinB=sin(A+B)sin(A-B)

Dear kiran,

Given the functions (sinα, cosα, sinβ and cos β), we seek a formula that expresses sin(α+β).

ABC which has an angle α
ACD which  "   "    "  β
The long side ("hypotenuse') of ACD is AD=R. Therefore

DC = R sin β
AC = R cos β
Similarly

BC = AC sin α = R cos β sin α
AB = AC cos α = R cos β cos α

The triangle ADF is right-angled and has the angle (α+β). Therefore

R sin (α+β) = DF
R cos (α+β) = AF

Start by deriving the sine:
R sin (α+β) = DF  =  EF + DE  =  BC + DE

 Note in the drawing the two head-to-head angles marked with double lines: like all such angles, they must be equal. Each of them is one of the two sharp ("acute") angles in its own right-angled triangle. Since the sharp angles in such a triangle add up to 90 degrees, the other two sharp angles must be equal. This justifies marking the angle near D as α, as drawn in the figure.
In the right-angled triangle CED

DE = DC cos α = R sin β cos α
EC = DC sin α = R sin β sin α
Earlier it was already shown that
BC = R cos β sin α
AB = R cos β cos α
Therefore
R sin (α+β)  =  BC+DE  =  R cos β sin α + R sin β cos α

Cancelling R we have

  sin (α+β)  =  sin α cos β  + cos α sin β

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