Question

4)sin^2a+sin^2b+ cos(a+b)*cos(a-b)=a

Answer 1

sin²α+sin²β+1/2*(2cos((2α+2β)/2)*cos((2α+2β)/2))=sin²α+sin²β+1/2(cos2α+cos2β)=sin²α+sin²β+1/2(1-2sin²α+1-2sin²β)=sin²α+sin²β+1/2(2-2sin²α-2sin²β)=sin²α+sin²β+1-sin²α-sin²β=1