F.4 a math 問題,急!!

we can prove it by simplifying the more complicated side
- sin(x+y)sin(x-y)= - (sinxcosy+cosxsiny)(sinxcosy-cosxsiny) (from the formular given)
= - [(sinxcosy)^2-(cosxsiny)^2] ( (a+b)(a-b)=a^2-b^2 )
= - (sin^2 x *cos^2 y - cos^2 x sin^2y)
= - (1-cos^2 x)cos^2 y - (1-cos^2 y)cos^2 x) (sin^2 x+cos^2 x=1)
= - (cos^2 y-cos^2 xcos^2 y-cos^2 x + cos^2 xcos^2 y)
= - (cos^2 y -cos^2 x)
= cos^2 x-cos^2y

cos^2 x - cos^2 y= - sin(x+y) sin(x-y)
LHS=COS^2x-COS^2y
=-sin((2X+2Y)/2)sin((2X-2Y)/2)
=-SIN(2(X+Y)/2)SIN(2(X-Y)/2)
=- sin(x+y) sin(x-y)
=RHS
IT IS IDENTITY