高一数学在线等急

显示全部楼层 · 2011-12-31 21:06:45

(x/2+π/4)=π/2+(x/2-π/4)，所以tan(x/2+π/4)tan(x/2-π/4)=-1f(x)=2√2cosx/2sin(x/2+π/4)-1=2sinx/cosx/2+2(cosx/2)^2-1=sinx+cosx=√2sin(x+π/4)f(x)的最大值是√2，最小值是-√2。0<=x<=π，则π/4<=x+π/4<=5π/4。π/4<=x+π/4<=π/2，则0<=x<=π/4。π/2<=x+π/4<=π，则π/4<=x<=3π/4。π<=x+π/4<=5π/4，则3π/4<=x<=π。所以，f(x)在区间[0,π/4]和[3π/4,π]上单调递增；在区间[π/4,3π/4]上单调递减。...

千问 · 2011-12-31 21:06:45

√2sin(x/2+π/4)=sinX/2+cosX/2tan(x/2+π/4)*tan(x/2-π/4)=-1(你自己拆开)f(x)=a·b=2cosx/2*√2sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)=2cosX+1-1=2cosx所以f(x)的最大值2，最小正周期为2π.f(x)在[0，π]上的单调区间为...

高一数学 在线等 急

高一数学在线等急