y=（sinx+cosx+1)/(cosx-1) 求函数的值域

显示全部楼层 · 2013-5-1 19:31:25

sinx=2sin(x/2)cos(x/2)
cosx=2cos^2(x/2)-1=1-2sin^2(x/2)带入得 y=(2sin(x/2)cos(x/2)+2cos^2(x/2))/(-2sin^2(x/2))
=-cot(x/2)-cot^2(x/2)令cot^2(x/2)=t 则 y=-t^2-t t为实数该函数为一元二次函数则其值域为 {-∞,1/4}...

千问 · 2013-5-1 19:31:25

首先函数式可以用二倍角公式化简：由sinx=2sin(x/2)cos(x/2);cosx+1=2[cos(x/2)]^2; cosx-1=-2[sin(x/2)]^2 带入化简得y=-1/tan(x/2)-1/[tan(x/2)]^2再令 t=1/tan(x/2)则y=-t-t^2=-(t+1/2)^2+1/4≤1/4...