Question

求证sin(a+b)sin(a-b)/sin'2acos'2b=1-tan'2b/tan'2a

Answer 1

sin(a+b)sin(a-b)=[sinacosb+cosasinb]*[sinacosb-cosasinb]=(sinacosb)²-(cosasinb)²
sin(a+b)sin(a-b)/sin'2acos'2b=[(sinacosb)²-(cosasinb)²]/sin'2acos'2b
=sin²acos²b(1-tan²b/tan²a)/sin²acos²b=1-tan'2b/tan'2a

Answer 2

sin(a+b)sin(a-b)/sin'2acos'2b
=(sinacosb+sinbcosa)(sinacosb-sinbcosa)/sin'2acos'2b
=[(sinacosb)^2-(sinbcosa)^2]/(sinacosb)^2
=1-(sinbcosa)^2/(sinacosb)^2
=1-(sinb/cosb)^2(cosa/sina)^2
=1-tan^2bcot^2a
=1-tan^b/tan^2a

追问

[(sinacosb)^2-(sinbcosa)^2]为什么等于1-(sinbcosa)^2

追答

[(sinacosb)^2-(sinbcosa)^2]/(sinacosb)^2
去括号后分别除
=(sinacosb)^2/(sinacosb)^2-(sinbcosa)^2/(sinacosb)^2
=1-(sinbcosa)^2/(sinacosb)^2