How do you prove sin(A+B) * sin(A-B) = sin2 A - sin2 B?

To Prove

sin(A+B) * sin(A-B) = sin2 A – sin2 B

Poof

We know that formula for sin(A+B) = sin(A+B)=sin(A)cos(B)+cos(A)sin(B)

Also sin(−B)=−sin(B)

cos(−B)=cos(B), so

sin(A−B)=sin(A)cos(B)−cos(A)sin(B)

Therefore sin(A+B)⋅sin(A−B)

=(sinAcosB+cosAsinB)(sinAcosB−cosAsinB)

=(sinAcosB)2−(cosAsinB)2

Now will use the identity (a+b)(a−b)=a2 – b2 in the above equation

=sin2Acos2B−sin2Bcos2A

=sin2A(1−sin2B)−sin2B(1−sin2A)

Now we know that  sin2θ+cos2θ=1 ( By Pythagoras theorem)

=sin2A−sin2B−sin2A sin2B+sin2Bsin2A

=sin2A−sin2B

sin(A+B) sin(A-B) = sin2 A – sin2 B

Leave a Comment

Your email address will not be published. Required fields are marked *

BOOK

Free Class