y=1 2sin(1 2x π 6)的对称中心

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函数y=sin(3x+π/12)sin(3x-5π/12)的最小正周期为

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

求函数y = sin(x+π/6)-cos(x+π/3) 的最大值和最小值

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

函数y=sin^2x+2/sinx,x∈[π/6,2π]的最小值

转换一下,令sinx=t再做,答案为-1

已知函数y=2sin(2x-π/3) 当x属于[π/12,π/3]时,y的取值范围?此函数由y

解由x属于[π/12,π/3]即π/12≤x≤π/3即π/6≤2x≤2π/3即-π/6≤2x-π/3≤π/3即-1/2≤sin(2x-π/3)≤√3/2即-1≤2sin(2x-π/3)≤√3即-1≤y

y=2sin(2x+π/6)得到函数y=g(x)求函数y=g(x)在区间【0,π/12】上的最大值

因0≤x≤π/12所以π/6≤2x+π/6≤π/3则y=2sin(2x+π/6)在[0,π/12]上的最大值为2sin(π/3)=√3即y=g(x)在[0,π/12]上的最大值为√3

y=sinωx+cosωx的一条对称轴是x=-π/12,求ω

y=sinωx+cosωx=根号2*sin(ωx+π/4)其对称轴为ωx+π/4=π/2+kπ,ωx=π/4+kπ,带入x=-π/12,解得ω=-(3+12/k),k为整数

y=sin(x+pi/6)sin(x-pi/6)的最小正周期

y=(sinxcospi/6+cosxsinpi/6)(sinxcospi/6-cosxsinpi/6)=(sinxcospi/6)^2-(cosxsinpi/6)^2=3/4*(sinx)^2-1/

函数y=sin(x+π/2)cos(x+π/6)的单调递减区间是?

y=sin(x+π/2)cos(x+π/6)=cosx*cos(x+π/6)=cosxcosx1/2根号3+1/2cosxsinx=1/2根号3cos^2x+1/4sin2x=1/2根号3*1/2(1

函数y=sin(2分之π+x)cos(6分之π-x)的最大值是

我来回答吧.看图片.我想这样你就可以看懂所有了.还有问题可以及时询问.

求函数y=sin^4x+cos^4x,x(0,π/6)的最小值

y=sin^4x+cos^4x=sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-1/2sin^2

求函数的值域y=sin(2x+π/3),x∈(-π/6,π)

∵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】

函数y=sin²(x+π/12)+cos²(x-π/12)-1的周期T=?,奇偶性为

答案:T=π奇函数y=1-(cos(x+π/12))^2+(cos(x-π/12))^2-1=(cos(x-π/12))^2-(cos(x+π/12))^2=(cos(x-π/12)-cos(x+π/

x*y'*sin(y/x)-y*sin(y/x)+x=0 求微分方程的解

y'sin(y/x)-y/x*sin(y/x)+1=0令y/x=u,则y'=u+xu'所以(u+xu')sinu-usinu+1=0xu'sinu+1=0-sinudu=dx/x两边积分:cosu=l

函数y=sin(x+π6

∵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

y=sin(3x-π/6)的导数

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~

求函数y=2sin(2x+π/6)+2sin(2x-π/12)的最大值,并求出此时x的集合.

令t=2x-π/12,则2x+π/6=2t+π/4,所以y=2sin(2x+π/6)+2sin(2x-π/12)=2sin(t+π/4)+2sint=√2sint+√2cost+2sint=(√2+2

将函数y=sinx的图像转变为y=sin(2x+π/6)

对的,还可以:1.将x坐标向左平移π/6个单位2.将x坐标扩大两倍,y坐标不变

求函数y=sin(x+派/6)sin(x-派/6)+acosx的最大值

y=sin(x+π/6)sin(x-π/6)+acosx=(3/4)(sinx)^2-(1/4)(cosx)^2+acosx=-(cosx)^2+acosx+3/4=-(cosx-a/2)^2+a^2

求函数y=sin(x+π/6)sin(x-π/6)+acos的最大值.(其中a为定值)

y=sin(x+π/6)sin(x-π/6)+acosx=-1/2[cos(x+π/6+x-π/6)-cos(x+π/6-x+π/6)+acosx=-1/2(cos2x-cosπ/3)+acosx=-

求函数y=sin²(x+π/12)+cos²(x-π/12)-1的最大值

y=sin²(x+π/12)+cos²(x-π/12)-1=(1-cos(2x+π/6))/2+(1+cos(2x-π/6))/2-1=1/2[-cos(2x+π/6)+cos(2