已知arcsin（sinα=+sinβ)+arcsin(sinα-sinβ)=kπ/2,k为奇数，求sin^2α+sin^2β的值

设A=arcsin（sinα+sinβ)，B=arcsin(sinα-sinβ)
故A+B=kπ/2
设k=2n+1(n∈Z)
故A+B=nπ+π/2
故cos(A+B)=cos(nπ+π/2)=0
sinα+sinβ=sinA
sinα-sinβ=sinB
解之：sinα=(sinA+sinB)/2
sinβ=(sinA-sinB)/2
故sin²α+sin²β
=1/2(sin²A+sin²B)
=1/4(1-cos2A+1-cos2B)
=1/2-1/4(cos2A+cos2B)
=1/2-1/2cos(A+B)cos(A-B)
=1/2-1/2cos(nπ+π/2)cos(A-B)
=1/2