3sinπx-log1 2x

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∫√(sin^3 x-sin^5 x)dx 上限π 下限0 求定积分

sin³x-sin^5x=sin³x(1-sin²x)=sin³xcos²x当00√(sin³xcos²x)=sinxcosx√s

化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1

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[

已知函数f(x)=2√3sin²x-sin(2x-π/3)

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间:π/2+2

函数f(x)=sin(x+π3)sin(x+π2)

y=sin(x+π3)sin(x+π2)=(sinxcosπ3+cosxsinπ3)cosx=12sinxcosx+32cos2x=14sin2x+32•1+cos2x2=34+12sin(2x+π3

已知函数f(x)= [sin(2π-x)sin(π+x)cos(-x-π)] /[2cos(π-x)sin(3π-x)]

∵f(x)=[-sinx(-sinx)cos(π+x)]/[-2cosxsin(π-x)]=[sin²x(-cosx)]/(-2cosxsinx)=1/2sinx∴最小正周期T=2π∴函数图

当x趋向π lim (sin 3x )/( sin 2x) 的极限为?

令a=π-x则a趋于0sin3x=sin(3π-3a)=sin3asin2x=sin(2π-2a)=-sin2a所以原式=-lim(a→0)sin3a/sin2asin3a和sin2a的等价无穷小是3

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x)

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

已知函数f(x)=2根号3sin平方x-sin(2x-π/3)

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

高中数学:已知函数f(x)=2sin(x+π/2).sin(x+7π/3)-

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3(cos*2x-sin*2x)=sin2x+根号3cos2x=2sin(2x+派

sin(x+π) sin(x+3/2π)怎么化简?

sin(x+π)=-ainx用诱导公式:x+π是第三象限的角,第三象限的角的正弦为负,π是1/2π的偶数倍,不改变函数名sin(x+3/2π)=-cosx用诱导公式:x+3/2π是第四象限的角,第四象

设f(x)为奇函数,且当x>0时,f(x)=log12x

(Ⅰ)设x<0时,则−x>0⇒f(−x)=log12(−x)⇒f(x)=−f(−x)=−log12(−x).所以:当x<0时,f(x)=-log 12(-x).(Ⅱ)由题意,得x>0log1

已知函数f(x)=sin^2 x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4),x属于R,求f(x

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

sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)

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

(2011•黑龙江一模)函数f(x)=3sinπ2x−log12x的零点个数是(  )

因为函数f(x)=3sinπ2x−log12x的零点个数就是对应的函数y=3sinπ2x与y=log 12x的交点个数.又因为函数y=3sinπ2x的周期T=2ππ2=4.而y=log&nb

化简sin(2x+π/3)-(根号3)sin²x+sinxcosx+(根号3)/2

∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3)

函数y=sin(x+π3

由题意x∈[0,π2],得x+π3∈[π3,5π6],∴sin(x+π3)∈[12,1]∴函数y=sin(x+π3)在区间[0,π2]的最小值为12故答案为12

已知函数f(x)=sin²ωx+根号3sinωx乘sin(ωx+π/2)+2cos²ωx,x∈R,(

(1)化解函数:√3∵f(x)=sin²wx+√3sinwxcoswx+2cos²wx=√3/2sin2wx+sin²wx+cos²wx+cos²wx