Need help with a couple of trigonometric identities?

Question

Hi,
Prove that
sin(x+y)sin(x-y) = sin^2x - sin^2y

Also Prove that
cos(x+y)cos(x-y) = cos^2y - sin^2x

Thanks in advance

Answer 1

1) (sinxcosy+sinycosx)(sinxcosy-sinycosx)
=(sinxcosy)^2-(sinycosx)^2
= sin^2x(1-sin^2y)-sin^2y(1-sin^2x)
= sin^2x-sin^2y
2) (cosxcosy-sinxsiny)(cosxcosy+sinxsiny)
=(cosxcosy)^2-(sinxsiny)^2
=cos^2y(1-sin^2x)-sin^2x(1-cos^2y)
=cos^2y-sin^2x

Answer 2

sin(x+y) sin(x-y)
= (SinxCosy + CosxSiny)(SinxCosy - CosxSiny)
= (Sinx)^2 (Cosy)^2 - (Cosx)^2(Siny)^2
= (Sinx)^2 [1 - (Siny)^2] - {[1 - (Sinx)^2](Siny)^2

multiply and cancel similar terms

= sin^2x - sin^2y

2nd part same but you have to use following formulae
cos(x+y) = cosx cosy - sinx siny
cos(x-y) = cosx cosy + sinx siny