Question

Prove that sin (A + B).sin (A-B) = sin^2A-sin^2B = cos^2B-cos^2A​

Answer 1

Answer:

$sin(A + B).sin(A - B) = sin^{2}A - sin^{2}B$

$sin(A + B).sin(A - B) = cos^{2}B - cos^{2}A$

Step-by-step explanation:

We have-

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

We know that-

sin(A + B) = sinAcosB + cosAsinB

sin(A - B) = sinAcosB - cosAsinB

Then sin(A + B).sin(A - B) =  (sinAcosB + cosAsinB).(sinAcosB - cosAsinB)

We know that-

$(a + b).(a - b) = a^{2} - b^{2}$

So-

sin(A + B).sin(A - B) =  (sinAcosB + cosAsinB).(sinAcosB - cosAsinB)

                               = {$sin^{2}A.cos^{2}B} - {cos^{2}A.sin^{2}B$}

But $sinA = 1 -cos^{2}A$

And $sinB = 1 - cos^{2}B$

sin(A + B).sin(A - B) = $(1 - cos^{2}A).cos^{2}B  - cos^{2}A(1 - sin^{2}B)$

          = $(cos^{2}B-cos^{2}A.cos^{2}B) - (cos^{2}A - cos^{2}.cos^{2}B)$

          = $cos^{2}B - cos^{2}A.cos^{2}B - cos^{2}A + cos^{2}A.cos^{2}B$$sin(A + B).sin(A - B) = cos^{2}B - cos^{2}A$

                                                                                       Proved.

Again, But-

$cos^{2}B = 1 - sin^{2}B$ and

$cos^{2}A = 1 - sin^{2}A$

So,  sin(A + B).sin(A - B) = $cos^{2}B - cos^{2}A$

                                       = $(1 - sin^{2}B) - (1 - sin^{2}A)$

                                       = $1 - sin^{2}B - 1 + sin^{2}A$

$sin(A + B).sin(A - B) = sin^{2}A - sin^{2}B$

                                                                                     Proved.