y=3cos(2 3x 5 6π)-2
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∵令x2−π3∈[-π+2kπ,2kπ],(k∈Z)可得x∈[-4π3+4kπ,2π3+4kπ],(k∈Z)∴函数y=cos(x2−π3)的单调递增区间是[-4π3+4kπ,2π3+4kπ],(k∈Z
y=12[1+cos2(x-π12]+12[1-cos2(x+π12]-1=12[cos(2x-π6)-cos(2x+π6)]=sinπ6•sinx=12sinx.T=π.故答案为:π.
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左加又减:y=cos(x-π/2)=sinx.因此只需要把图象向右平移5π/6,即cos(x-5π/6+π/3)=cos(x-π/2)
这个函数应该是y=cos(πx/3+φ)吧?少了一个x,由πx/3+φ)=kπ,将x=9π/4代入得到φ=-3π/4+kπ,令k=1得φ=π/4,所以函数y=sin(2x-φ)的增区间由不等式-π/2
cos(x+π/3)=sin[π/2-(x+π/3)]=sin(π/6-x)=-sin(x-π/6)所以y=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)=si
用-x代入可得左边括号为-x+π/3因为cos是偶函数所以左边括号等于π/3-x;右边一个括号里面刚好是-x-π/3同理知道等于x+π/3所以相当于左右两个换了一下顺序所以为偶函数
由三角函数的周期公式,可得T=2π25=5π,即函数的最小正周期为5π故答案为:5π
∵y=cos(π6−x)=cos(x-π6),由2kπ-π≤x-π6≤2kπ,k∈Z得:2kπ-56π≤x≤2kπ+π6,k∈Z.∴原函数的单调递增区间为[2kπ-56π,2kπ+π6](k∈Z).故
∵y=cosx+cos(x-π3)=cosx+cosxcosπ3+sinxsinπ3=32cosx+32sinx=3(cosπ6cosx+sinπ6sinx)=3cos(x-π6),∵-1≤cos(x
y=cos(π/3-x)y'=-sin(π/3-x)*(-1)=sin(π/3-x)y=e^3xy'=e^(3x)*3=3e^(3x)y=In(3-x)y'=1/(3-x)*(-1)=1/(x-3)y
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
y=cosx+cos(x+π/3)=cosx+cosxcos(π/3)-sinxsin(π/3)=3cosx/2-√3sinx/2=√3(sin(π/3)cosx-cos(π/3)sinx)=√3si
y=cosx-√3sinx=2(1/2cosx-√3/2sinx)=2(cosxcosπ/3-sinxsinπ/3)=2cos(x+π/3)再问:那也可以化成2sin(x-π/6)吗再答:y=cosx
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
由y=cosx的图象先向左平移π3个单位,再把各点的纵坐标不变,横坐标变为原来的13倍,即可得到y=cos(3x+π3)的图象.故答案为:左;π3;缩小;13.
y=cos(x-π/3)=cosx*cosπ/3+sinxsinπ/3y=cos(-x-π/3)=cos-x*cosπ/3+sin-xsinπ/3=cosx*cosπ/3-sinxsinπ/3非奇非偶
f(π/3)=f(-π/3)偶函数!再问:要证明啊这种办法只能用来验证是否是吧。。。。求证明的过程再答:f(a)=cos(π/3-a)cos(π/3+a)f(-a)=cos(π/3+a)cos(π/3
由x-π3∈[2kπ,2kπ+π],可得x∈[π3+2kπ , 4π3+2kπ](k∈Z),∴函数y=cos(x-π3)的单调递减区间是[π3+2kπ , 4π
1.两边求导得:y'=-sin(x-y)(1-y')解得y'=sin(x-y)/[sin(x-y)-1]2.y'=-e^-xy''=e^-xy'"=-e^-x3.y'"=(e^2x)'"(sinx)+