Asked by saurav

show that :
sin(A+B).sin(A-B)=Cos^2B - Cos^2A =sin^2A - sin^2B
Answered by Reiny
let's just tackle the LS

LS = sin(A+B)*sin(A-B)
= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)
= sin^2 Acos^2B - cos^2Asin^2B , the difference of squares pattern
= <b>(1 - cos^2A)(cos^2B) - cos^2A(1-cos^2B)</b>
= cos^2B - cos^2Acos^2b - cos^2A + cos^2Acos^2B
= cos^2B - cos^2A
= RS

In the line in bold, had I replaced cos^2 x with 1 - sin^2 x , I would obtain the second version of the Right Side

Answered by saurav
thank you . but what about the 3rd part of the question ? sin^2A - sin^2B what do we need to do with it?
Answered by Reiny
Did you not read my post?
I told you how to do it.
Answered by saurav
oh yeas. sorry for the turmoil
There are no AI answers yet. The ability to request AI answers is coming soon!