作业帮根号cosx-cos^3x原函数
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sin^3X+cos^3X=（sinx+cosx)(sin^2x-sinxcosx+cos^2x)=根号2(1+sinxcosx)sinX+cosX=根号2所以sin^2x+cos^2x+2sinxc
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
先求积分再代入范围.原式=∫√cosx[1-(cosx)^2]dx=∫|sinx|√(cosx)dx=-2∫√(cosx)d(cosx)此处积分范围变为(0,π/2)=-2*(2/3)(cosx)^(
因为cos2x=2cos^2x-1;sin2x=2(sinx)(cosx)所以f(x)=cos^2x+(根号3)sinx*cosx=(cos2x+1)/2+[(根号3)(sin2x)]/2因为1/2=
因为cos2x=2cos^2x-1;sin2x=2(sinx)(cosx)所以f(x)=cos^2x+(根号3)sinx*cosx=(cos2x+1)/2+[(根号3)(sin2x)]/2因为1/2=
sinx+cosx=√2[（√2/2）sinx+(√2/2)cosx]=√2sin(x+π/4)=√2cos[π/2-(x+π/4)]=√2cos(π/4-x)或=√2cos(x-π/4)
√3sinx-cosx+2cos(x-π/6)=2sin(x-π/6)+2cos(x-π/6)=2√2sin[(x-π/6)+π/4]=2√2sin(x+π/12)
（1）由f（x）=cosx+根号3cos（x+二分之兀）化简得：f(X)=-2sin(x-π/6)要f(X)有最大值,则sin(x-π/6)=-1故：X-π/6=-π/2+2Kπ,K∈Z得出X=-π/
1.f(x)=sinx/3cosx/3+根号3cos^2x/3=1/2sin2x/3+根号3/2(cos2x/3+1)=sin(2x/3+pi/3)+根号3/2对称中心的横坐标满足sin(2x/3+p
cosx+根号3sinx=2*[1/2cosx+(根号3)/2sinx]=2*【sinπ/3*2cosx+cosπ/3*sinx】=2cos(x-π/3)希望我的答案您能满意,
f(x)=cos^2x+√3sinx*cosx=(cos2x+1)/2+√3/2sin2x=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+1/2所以当2x+π/6=2kπ+π/
1、根号下大于等于0log2(4-x)>=0=log2(2)所以4-x>=2x
cosx+cos(x-派/3)=2cos(2x-π/3)/2cos(x-x+π/3)/2=2cos(x-π/6)×cosπ/6=2×（-√3/3）×√3/2=-1
f(x)=1/2sinx/2+√3/2(1+cosx/2)=1/2sinx/2+√3/2cosx/2+√3/2=sin(x/2+π/6)+√3/2
y=cosx-√3sinx=2(1/2cosx-√3/2sinx)=2(cosxcosπ/3-sinxsinπ/3)=2cos(x+π/3)再问：那也可以化成2sin(x-π/6）吗再答：y=cosx
∵sin(x/2)=±√[(1-cosx)/2]cos(x/2)=±√[(1+cosx)/2又∵x∈（3π/2,2π）∴3π/4
原式＝cosx-根号3cos30cosx+根号3sin30sinx+cosxcos60-sinxsin60=cosx-3/2cosx+根号3/2sinx+1/2cosx-根号3/2sinx=0