Thread: compound-angle formulae with trig identity question

sin(A+B)sin(A-B) = sin^2A-sin^2B is what I have to prove

So
LHS= (sinAcosB+cosAsinB)(sinAcosB-cosAsinB)

This is obviously a quadratic. I just don't know how it would multiply

e.g sinAcosB x sinAcosB = ???

==>sin^2A.cos^2B-sin^2Bcos^2A
===> sin^2A(1-sin^2B)-sin^2B(1-sin^2A)
===> sin^2A-sin^2Asin^2 B-sin^2B+sin^2Asin^2B
===> sin^2A-sin^2B.

sin(A+B)=sinAcosB+cosAsinB
sin(x) = cos(90- x) and cos(x) = sin(90 - x)
let x = A + B,
we have,
sin(A+B)=cos (90-(A+B))
then i can write this
cos90-A-B)as cos((90-A)-B)
cos(A-B) =cos(A)cos(B) + sin(A)sin(B)
sin(A+B)=cos(90-A)cosB +sin(90-A)sin(B)
sin(A+B)=sinAcosB+cosAsinB
i think these prooves can help you...

Thankyou

you are welcome