y=1 2sin(3x-π 6)五点法
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![y=1 2sin(3x-π 6)五点法](/uploads/image/f/907955-35-5.jpg?t=y%3D1+2sin%283x-%CF%80+6%29%E4%BA%94%E7%82%B9%E6%B3%95)
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
振幅为2;周期为π;初相为π/3单增区间:kπ-5π/12≦x≦kπ+π/12对称轴:x=﹙1/2﹚kπ+(1/12)π
你把括号里的看成一个整体记作t.这样自变量是t,就是y=sint的简单正弦函数,不同的t对应求出不同的x即可
y=sin(3x+π/12)sin(3x-5π/12)=sin(π/2-3x-π/12)sin(3x-5π/12)=cos(5π/12-3x)sin(3x-5π/12)=cos(3x-5π/12)si
y=sinxcos30+cosxsin30-cosxsin60-sinxcos60=sinx[(根号3-1)/2]+cosx[(1-根号3)/2]=[(根号3-1)/2](sinx-cosx)=[(根
y=(1/2)[1-cos(4x+2π/3)]y'=2*sin(4x+2π/3)
y=cos^2(3x+π/6)-sin^2(3x+π/6)=cos[2(3x+π/6)]=cos(6x+π/3)
y=sinx先向左平移π/6,得y=sin(x+π/6)然后,纵坐标不变,横坐标变为原来的1/2,得y=sin(2x+π/6)最后,向上平移3/2,得y=sin(2x+π/6)+3/2
我列个去,就算我高中毕业到现在已经8年了,我也看的出来1楼的乱说的撒,值域明显是[-2,2]嘛
∵x∈(-π/6,π); ∴2x+π/3∈(0,2π+π/3); 则函数y的最大值为1,最小值为-1; 则y∈【-1,1】
y=(√3/2)sin(x+π/2)+cos(π/6-x)=(√3/2)cosx+cos(π/6)cosx+sin(π/6)sinx=(√3/2)cosx+(√3/2)cosx+(1/2)sinx=√
∵π3≤x≤3π4∴π3≤2x−3π4≤7π6,根据正弦函数图象则−12≤sin(2x−π3) ≤1,故答案为[−32,3].
∵0≤x≤π2,∴π6≤x+π6≤2π3;∴当x+π6=π2时,函数取得最大值是y=sin(x+π6)=1;当x+π6=π6时,函数取得最小值是y=sin(x+π6)=12;∴函数y=sin(x+π6
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
通过复合函数求导,可以得到y'=cos(3x-π/6)*3=3cos(3x-π/6)欢迎追问~
y=sin(2x+π/3)+cos(2x-π/6)=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x=sin2x+√3cos2x=2sin(2x+π/3)2k
正在做啊再问:恩再答:cos[π/2-(π/3+x)]=cos(π/6-x)=sin(π/3+x)y=sin(x+π/3)cos(π/6-x)=sin(x+π/3)sin(π/3+x)=sin
任何正弦函数,只要系数是1,值域就是[-1,1]