y=3sin(2x-π╱4)简图
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因为由上式可知y=cos(3x+π\4)=-sin[3(x-π/12)],要将y=-sin[3(x-π/12)]变换到y=sin(-3x),则需要作加法,即:-sin[3(x-π/12+π/12)],
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
y=(1/2)[1-cos(4x+2π/3)]y'=2*sin(4x+2π/3)
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
sin(π/4+x/2)sin(π/4-x/2)=sin(π/4+x/2)sin[π/2-(π/4+x/2)]∵π/4=π/2-π/4∴sin(π/4-x/2)=sin(π/2-π/4-x)=sin[
∵-π6<x<π6,∴0<2x+π3<2π3,根据正弦函数的性质,则0<sin(2x+π3)≤1,∴0<2sin(2x+π3)≤2∴函数y=2sin(2x+π3) (-π6<x<π6)的值域
y=2sin[(1/2)x-π/3]+1(1)最小正周期:T=2π/(1/2)=4π.(2)单调性:由2kπ-π/2≦(1/2)x-π/3≦2kπ+π/2,得2kπ-π/6≦(1/2)x≦2kπ+5π
∵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].
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
∵函数表达式为y=3sin(2x+π4),∴ω=2,可得最小正周期T=|2πω|=|2π2|=π故答案为:π
sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/
答:y=sin(2x+3兀/2)y=sin(2x+2兀-1/2*兀)y=sin(2x-兀/2)y=-sin(兀/2-2x)y=-cos(2x)y=-(cosx)^2+(sinx)^2所以f(-x)=-
令2kπ+π2≤3x+π4≤2kπ+3π2,k∈z,求得2kπ3+π12≤x≤2kπ3+7π36,故函数的减区间为[2kπ3+π12,2kπ3+7π36],k∈Z,故答案为:[2kπ3+π12,2kπ
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/
1y=sinX向左移动π/4,得到y=sin(x+π/4)2y=sin(x+π/4)沿x轴压缩为原来的1/3,得到y=sin(3x+π/4)3y=sin(3x+π/4)沿y轴扩大2倍,得到y=2sin