设π/3<x<3π/4,Sin(x-π/4)=3/5,求（Sinx-cos2x+1）/tanx的值

显示全部楼层 · 2013-3-2 00:34:37

解：sinx-cos2x+1/tanx=sinx-cosx^2+sina^2+cosx^2+sina^2/sinx/cosx
=1+2sinx/cosx
由sin(x-π/4)=3/5 嘚sinx*√2/2-cosx*√2/2=3/5 即sinx=cosx+3√2/5······①
sinx^2+cosx^2=1·······②
联立①②解得cosx=√2/10或-7√2/10(舍)
∴sinx=7√2/10
∴1+2sinx/cosx=1+7√2/5/√2/10=14+5√2
即原式（Sinx-cos2x+1）/tanx的值为1...