f(x)=根号3sinx-cosx用五点法
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原式可化为:√3sinxcosx-sin²x=(√3/2)sin2x+(cos2x)/2-1/2=sin(2x+π/6)-1/2故最小正周期为2π/2=π
已知:函数f(x)=2sinxcosx+2√3cos²x-√3求:(1)单调增区间和最小正周期;(2)当x∈[-π/4,π/4]时求最值.f(x)=2sinxcosx+2√3cos²
F(x)=m.n=2根号3sinxcosx+cos^2x-sin^2x=根号3sin2x+cos2x=2sin(2x+π/6)
f(x)=2(sin兀/3cosx-cos兀/3sinx)=2sin(兀/3-x)0
f(x)=(sinx-1)/根号(3-2cosx-2sinx)=-(1-sinx)/根号[(sin²x-2sinx+1)+(cos²x-2cosx+1)]=-(1-sinx)/根号
y=(sinx+根号3)/cosxycosx=sinx+√3sinx-ycosx=-√3√(1+y²)sin(x-β)=-√3sin(x-β)=-√3/√(1+y²)∵|sin(x
f(x)=向量m.向量nf(x)=2sin^2x+2√3sinxcosx.=1-cos2x+√3sin2x.∴f(x)=2sin(2x-π/6)+1.(1)函数f(x)的最小正周期T:T=2π/2,∴
f(x)=sinx+根号3cosx=2*sin(x+pi/3)1.T=2pi2.x用x-pi/3代替:y=sinx单调增区间:【0,pi/2】
(1)f(x)=msinxcosx-2√3(sinx)^2+√3=(m/2)sin2x+√3cos2x,x=π/6是函数y=f(x)的零点,∴(m/2+1)√3/2=0,m=-2.∴f(x)=-sin
万能代换:设sinx=2k/(1+k^2),cosx=(1-k^2)/(1+k^2).代入得y=(3k^2+2k+3)/√(5k^2+5+8k+3-3k^2)=(3k^2+2k+3)/√2(k^2+4
画图:∴x∈[0,π/2]时 值域为[-2,2]
设y=√2sinx+cosx,求导得:y‘=√2cosx-sinx,当y‘=√2cosx-sinx>0时,y=√2sinx+cosx为增函数,tanx0,cosx√2时,为减函数;当sinx=√6/3
f(x)=sinx+√3cosx=2(1/2sinx+√3/2cosx)=2(cosπ/3sinx+sinπ/3cosx)=2sin(x+π/3)所以最小正周期为:2π振幅为2再答:请采纳哦,谢谢再答
等于2sin(2x-六分之派)T=派
f(x)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)-1≤sin(x-π/6)≤1-2≤f(x)≤2值域是[-2,2]
f(x)=2(sinx*√3/2-cosx*1/2)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)、1/2表示2分之1=====sinx=4/5,题目应该有x∈〔-π/2,
y=2(sinx*1/2+cosx*√3/2)=2(sinxcosπ/3+cosxsinπ/3)=2sin(x+π/3)-π/6
f(x)=sinx+根号3cosx=2sin(x+π/3),即最小正周期为2π得到的g(x)=2sin(x+π/3-π/3)=2sinx,即在(O,π/2】上单调递增,在【π/2,π)上单调递减
1、由x范围则cosx>0sin²x+cos²x=1所以cosx=3/5所以f(x)=(4√3-3)/52、f(x)=2(sinx*√3/2-cosx*1/2)=2(sinx
f(x)=√3sinx-cosx=2(√3/2sinx-1/2cosx)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)所以最小值是-2