Question

已知sin（α+β）sin（α-β）=1/5，求sin²α-sin²β的值.

Answer 1

sin（α+β）sin（α-β）=1/5
(sinαcosβ+cosαsinβ）(sinαcosβ-cosαsinβ）=1/5
sin²αcos²β-cos²αsin²β=1/5
sin²α（1-sin²β)-(1-sin²α)sin²β=1/5
所以
sin²α-sin²αsin²β-sin²β+sin²αsin²β=1/5
即
sin²α-sin²β=1/5

Answer 2

(sina)^2-(sinb)^2
=(1-cos2a)/2-(1-cos2b)/2
=1/2*(cos2b-cos2a)
=1/2*{cos[(a+b)-(a-b)]-cos[(a+b)+(a-b)]}
=1/2*[cos(a+b)cos(a-b)+sin(a+b)sin(a-b)-cos(a+b)cos(a-b)+sin(a+b)sin(a-b)]
=1/2*2sin(a+b)sin(a-b)
=sin(a+b)sin(a-b)
=1/5 .