Question

what us the answer for sin^2A - sin^2B​

Answer 1

Answer:

We know that cos 2A=1–2sin^2A

or 2 sin^2A=1-cos 2A

sin^2A+sin^2B=(1/2)[2sin ^2A+sin ^2B]

=(1/2)[1-cos 2A+1-cos 2B]

=(1/2)[2-{cos 2A+cos 2B}]

=(1/2)[2–2cos(A+B).cos(A-B)]

=1-cos (A+B).cos (A-B) , answer

Answer 2

Answer:

sin^2A-sin^2B=sin(A+B).sin(A-B)

Step-by-step explanation:

1)it is a formula

2)sin(A+B)sin(A-B)

   (sinAcosB+sinBcosA)(sinAcosB-sinBcosA)

   sin^2(A).cos^2(B) - sin^2(B).cos^2(A)

                                           since cos^2A is equal to 1-sin^2A

  sin^2(A) - sin^2(A).sin^2(B)- sin^2(B) + sin^2(B).sin^2(A)

  sin^2(A) - sin^2(B)

      hence proved//