试求y=3sinπ 3x-4cosπ3 3x
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函数的周期T=2πω=2π2=π,由-π2+2kπ≤2x+π3≤π2+2kπ,解得−5π12+kπ≤x≤π12+kπ,即函数的递增区间为[−5π12+kπ,π12+kπ],k∈Z,由2x+π3=π2+
∵y=sin(2x+π3),∴由2kπ−π2≤2x+π3≤2kπ+π2,k∈Z.得kπ-5π12≤x≤kπ+π12,k∈Z.∴当k=0时,递增区间为[0,π12],当k=1时,递增区间为[7π12,π
π/2+2kπ再问:换元法有没有?再答:令3x+π/4=t,y=2sint的递减区间是:π/2+2kπ
y=-1/2sin(2/3x-π/4)所以y和sin(2/3x-π/4)单调性相反sinx的增区间是(2kπ-π/2,2kπ+π/2)减区间是(2kπ+π/2,2kπ+3π/2)所以sin(2/3x-
解题思路:三角函数图像解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.
y=3sin(3x+π/4)单调增区间是:2kPai-Pai/2
y=2sin(3x+π/4)依题意-π/2+2kπ
复合函数应该用链式法则求导:若y=g(u),u=f(x),则dy/dx=dy/du*du/dxy=sin^5xdy/dx=dsin^5x/dsinx*dsinx/dx=5sin⁴xcosx
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
y=6(2x+3)^2y=(e^x^2)2x-2y=cos(π/2x+4)×((-2π/(2x+4)^2))希望我写得清楚
sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c
y∈[1,3]当y=1时,sin(x+π/3)=-1,x+π/3=2kπ-π/2,x=kπ-5π/12,k∈Z当y=3时,sin(x+π/3)=1,x+π/3=2kπ+π/2,x=kπ+π/12,k∈
Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2
sinx的减区间是(2kπ+π/2,2kπ+3π/2)所以这里2kπ+π/2
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
1、y=(cos^2x+sin^2x)^2-2cos^2xsin^2x=1-1/2(sin2x)^2=1-1/4(1-cos4x)=3/4+1/4cos4x周期T=2pi/4=pi/22、y=(根3/
y=2sinxcos^2x/(1+sinx)=2sinx﹙1-sin²x﹚/(1+sinx)=2sinx﹙1-sinx﹚=-2﹙sinx-½﹚²+½y=sin^
y=(sinxcosπ/3-cosxsinπ/3)sinx=(1/2*sinx-√3/2*cosx)sinx=1/2*sin²x-√3/2*sinxcosx=1/2*(1-cos2x)/2-
y=sinx增区间[2kπ-π/2,2kπ+π/2]所以本题,2kπ-π/2≤π/4+2x≤2kπ+π/2kπ-3π/8
dy/d(x^3)=(dy/dx)/(d(x^3)/dx)=cosx/3(x^2)