作业帮fx=sin[四分之五π-x]-cos[四分之π x]
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/03 09:48:16
f(x_=(cosx+sinx)(cosx-sinx)=cos²x-sin²x=cos2x所以T=2π/2=πf(α/2)=cosα=1/3sin²α+cos²
f(x)=cosx-cos(x+π/2)=cosx+sinx=3/4sin^2x+cos^2x+2sinxcosx=9/162sinxcosx=sin2x=9/16-1=-7/16
四分之三X+四分之一=八分之五两边乘86x+2=56x=5-26x=3x=3/6x=1/2x-3/4x=181/4x=18x=18x4x=72
5/6x=5/6÷4/25x=5/6÷4/25÷5/6x=1÷4/25x=25/4x=6.25
(1)f（x）=sin（2x+π/6）+3/2,最小正周期为2π/2=π,单增区间为2Kπ-π/2
(1) 等式化简后：f(2)=±（√19／2）+3
fx=2cos²ωx+2sinωxcosωx+1＝1+cos2ωx+sin2ωx+1＝√2sin（2ωx+π/4)+2T=2π/ω,π/2=2π/2ω,ω=2f(x)=√2sin（x+π/4
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/
f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co
f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=(1/2)cos2x+(√3/2)sin2x+(cos(π/2)-cos2x)=-(1/2)cos2x+(√3/2)sin
T=2π/2=π[-1,1]最大值为1,最小值为-1
∵0＜a＜四分之π＜β＜四分之三π∴3π/4
4x/7=5(x-6)/835(x-6)=32x35x-35*6=32x3x=35*6x=70
你的分析前一半是对的,一直到“那么2x的单调增区间是[-4分之π,4分之π]”.2x的单调递增区间是[-π/2,π/2],x的才是[-π/4,π/4].所以函数在x=-π/3处取得最小值为-2分之根号
解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)
解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数
f(x)=sin(2x+π/6)-cos2x+1所以为2π/2=πf(x)=根号3/2sin2x-(cos2x)/2+1=sin(2x-π/6)+1所以最大值为2,x=π/2+2kπ-π/6=π/3+
f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2=1/2*sinx+√3/2*(1-cosx)+√3/2=1/2*sinx-√3/2*cosx+√3=sin(x