若tan(Л-α)=2,则2sin(3Л+ α)×cos(5Л/2+α)+sin(3Л/2-α)×sin(Л-α )的值为

问题描述:

若tan(Л-α)=2,则2sin(3Л+ α)×cos(5Л/2+α)+sin(3Л/2-α)×sin(Л-α )的值为

tan(Л-α)=2,则tana=-2,sina=-2cosa2sin(3Л+ α)×cos(5Л/2+α)+sin(3Л/2-α)×sin(Л-α )=2(-sina)*(-sina)+(-cosa)*sina=2(sina)^2-sinacosa=2-2(cosa)^2-(-2(cosa)^2)=2