求函数y=sinπ 3x cosπ 3x的减区间
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/05 22:58:20
sin2α+根号3倍cos2α=2sin(2α+π/3)=1sin(2α+π/3)=1/22α+π/3=5π/6α=π/4f(x)=2xcos(α-π/4)+sin(α+π/4)=2x+1a1=1a2
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
f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)=(1-cos2ωx)/2+(√3/2)sin2ωx=sin(2ωx-π/6)+1/2∵函数y=f(x)的图像相邻两条对称轴之间
cos2ωx=1-2sin²ωxf(x)=sin²ωx+√3cosωxcos(π/2-ωX)=(1-cos2ωx)/2+√3cosωxsinωx=1-(1/2)cos2ωx+(√3
1、(1)、y=√3/2sin2ωx-1/2cos2ωx+1=sin(2ωx-π/6)+1,T=2π/|2ω|=π,故|ω|=1,又当x=π/6时,函数有最小值,所以ω=-1.∴y=1-sin(2x+
(1)f(x)=√3sinωxcosωx-cos²ωx-1/2=√3/2*2sinωxcosωx-(2cos²ωx-1)/2-1=(√3/2)*sin(2ωx)-1/2*cos(2
∵(π3+4x)+(π6-4x)=π2,∴cos(4x-π6)=cos(π6-4x)=sin(π3+4x),∴原式就是y=2sin(4x+π3),这个函数的最小正周期为2π4,即T=π2.当-π2+2
y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=sin^2(2x)/2=(1-cos4x)/4周期显然是pi/2
y=sin^4x+cos^4x+4sin^2xcos^2x-1=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=si
∵[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0==>xcos(x+y)dx+xcos(x+y)dy+sin(x+y)dx=0==>xcos(x+y)(dx+dy)+sin(
f(x)=sin2ωx+√3cos2ωx=2sin(2ωx+π/3),两对称轴之间的最小值为π/2即半个周期,则周期为π=2π/2ω,所以w=1,所以f(x)=2sin(2x+π/3),f(α)=2s
f(x)=(1+cos2ωx)/2+√3sin2ωx=sin(2ωx+π/6)+1/2T=2π/(2ω)=π,ω=1,f(x)=sin(2x+π/6)+1/2(1)f(2π/3)=sin(4π/3+π
合并同类项么,很简单的只要你愿意去做左边=cos*x(cos*y+sin*y)+sin*x(cos*y+sin*y)=cos*x+sin*x=1=右边
原式=√3sinωxcosωx-cos^2ωx,其周期T=π/2.原式=2cosx[√3/2(sinωx-(1/2)cosωx]=2cosωx[sinωxcos(π/6)-cosωxsin(π/6)]
f(x)=√3sinωxcosωx+cosωx^2=√3/2*sin2ωx+1/2*(cos2ωx-1)=cosPi/6*sin2ωx+sinPi/6*cos2ωx-1/2=sin(2ωx+Pi/6)
y=sin⁴3xcos³4xdy/dx=cos³4x*d(sin⁴3x)/dx+sin⁴3x*d(cos³4x)/dx=cos
t=sinx+cosx=√2sin(x+π/4)-√2=再问:上面那个颠倒的V是什么再答:那是根号呀,√2表示根号2.再问:sin^2x这个颠倒的^也是根号?再答:这个是次方符号呀,sin^2x表示的
y=sinωxcosφ+cosωxsinφ=sin(ωx+φ).∵函数的最小正周期为π,∴ω=2,则y=sin(2x+φ).又x=π3是其图象的一条对称轴,∴2π3+φ=π2+kπ,φ=kπ−π6,k
(1)y=√3sinαxcosαx-cosαx^2+3/2=√3sin2αx/2-(cos2αx+1)/2+3/2=(√3sin2αx)/2-(cos2αx)/2+1=sin(2αx-30`)+1(2