lim ln(x 根号sinπx)
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1f(x)=√3sinπx+cosπx=2((√3/2)sinπx+(1/2)cosπx)=2sin(πx+π/3)∴最小正周期T=2π/w=2π/π=2值域f(x)∈[-2,2]2-π/2+2kπ<
利用三角函数积化和差公式sinAcosB=0.5[sin(A+B)+sin(A-B)]和倍角公式cos2A=cos²A-sin²A原式=sin²x+√3[sin2x+si
2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[
-2*cos根号x因为d(根号x)=1/(2*根号x)
由题意可得:f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2=2sin(2x+π/3
lim[ln(1+x)-lnx]/x=limln[(1+x)/x]/x=limln(1+1/x)/x=0.
∵cosπ/6=√3/2sinπ/6=1/2cosπ/6sinx+sinπ/6cosx=sin(x+π/6)∴√3sinX+cosX=2(√3/2sinX+1/2cosX)=2(cosπ/6sinx+
√2sin(x-π/4)=√2(sinxcosπ/4-cosxsinπ/4)=√2[(1/√2)sinx-(1/√2)cosx]=sinx-cosx
f(x)=2sinxcosx+2√3sin²x-√3=2sinxcosx+√3sin²x+√3(sin²x-1)=2sinxcosx+√3sin²x-√3cos
√[(sinx)^4+4(cosx)^2]-√[(cosx)^4+4(sinx)^2]=√[((sinx)^2-2)^2]-√[((cosx)^2-2)^2]=(sinx)^2-2-[(cosx)^2
f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T
前提掌握:sinx*sinx+cosx*cosx=1cos2x=2*cosx*cosx-1=1-2*sinx*sinxcos(x-π/4)=-sin(x-π/4+π/2)=-sin(x+π/4)sin
lim(x→0+)ln(sin3x)/ln(sinx)=lim(x→0+)[3cos3x/(sin3x)/[cosx/sinx]=lim(x→0+)(3sinx/sin3x=1再问:[3cos3x/(
f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x
原式=sin(x+π/3)+√3cos(x+π/3)+2sin(x-π/3)=2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3)=2sin(x+π/3+π/6)+2
∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3)
(1)化解函数:√3∵f(x)=sin²wx+√3sinwxcoswx+2cos²wx=√3/2sin2wx+sin²wx+cos²wx+cos²wx
通过泰勒公式可以在0点展开ln(x+√(1+x^2):ln(x+√(1+x^2)=x+o(x)o(x)表示余项是x的高阶无穷小所以代入原式=limln(x+√(1+x^2))/x=lim[x+o(x)