题目

tanα=m,π<α<2π,sinα=

提问时间：2020-12-03

答案

tanα=sinα/cosα=m
cosα=sinα/m
(sinα)^2+(cosα)^2=1
(sinα)^2+(sinα/m)^2=1
(m^2+1)/m^2*(sinα)^2=1
(sinα)^2=m^2/(m^2+1)
π<α<2π
sinα<0
sinα=-m/(m^2+1)^(1/2)