题目

tanα=2,求sin(π-α)cos(2π-α)sin(-α+ 3π/2)/tan(-α-π)sin(-π-α)的值

提问时间：2020-07-26

答案

sin(π-α)cos(2π-α)sin(-α+ 3π/2)/tan(-α-π)sin(-π-α)=sinαcosα(-cosα) / (-tanα)sinα=(cosα)^2 / tanα=(cosα)^2 / 2=[(1/2)(1+cos2α)] / 2=(1/4)(1+cos2α)=(1/4) + (1/4)×[1-(tanα)^2 / 1+(ta...