设函数y=cos(1 2x π 3),x属于(28π 5,a)
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y=2cos(x+π4)cos(x−π4)+3sin2x=2(12cos2x−12sin2x)+3sin2x=cos2x+3sin2x=2sin(2x+π6)∴函数y=2cos(x+π4)cos(x−
(1)f(x)=cos(x+2π/3)+2cos²(x/2)=-(cosx)/2-(√3sinx)/2+1+cosx=1-[(√3sinx)/2-(cosx)/2]=1-[sin(x-π/6
y=sin²x+2sinxcosx+3cos²x=1/2(1-cos2x)+sin2x+3/2(1+cos2x)=sin2x+cos2x+2=√2sin(2x+π/4)+2∵-π≤
(1)f(x)=cos(2x+π3)+sin2x=cos2xcosπ3−sin2xsinπ3+1−cos2x2=12−32sin2x所以函数f(x)的最大值为1+32,最小正周期π.(2)由f(x)=
利用sin(x+y)=sin(x)cos(y)+cos(x)sin(y)可知f(x)=2+sin(3x+π/12+x+π/6)=2+sin(4x+π/4)(1) 对称中心为((-π/4+kπ
(1)f(x)=cos2xcosπ3+sin2xsinπ3-cos2x-1=32sin2x-12cos2x-1=sin(2x-π6)-1…3分∴函数f(x)的最小正周期是T=2π2=π,…5分由2kπ
就是简单的复合函数求导问题嘛.1.y'=[1/cos(10+2x)]*[-sin(10+2x)]*2=[-2sin(10+2x)]/cos(10+2x)2.y'=[1/cos(3+x²)]*
利用诱导公式和三角恒定公式来解f(x)=1/2sin2x+√3cos^2=1/2sin2x+√3(1+cos2x)/2=1/2sin2x+√3/2*cos2x+√3/2=sin(2x+π/3)+√3/
求导得:f′(x)=-4sinxcosx+23cos2x=-2sin2x+23cos2x=4sin(π3-2x),令f′(x)=0,得到x=π6,∵f(0)=2+a,f(π2)=a,f(π6)=3+a
y=-2cos(1/2x+π/3)取极值时,x=2kπ+π/3且4π+π/3<28π/5
因为y=cos(3π2−x)cos(3π−x),所以结合诱导公式可得:y=tanx,所以根据正切函数的周期公式T=πω可得函数y=cos(3π2−x)cos(3π−x)的周期为:π.故答案为:π.
函数y=2sin(2x+π3)的图象关于点P(x0,0)成中心对称,所以2x+π3=kπ,k∈Z;所以x=kπ2−π6 k∈Z,因为x0∈[−π2,0],所以x0=−π6;故答案
由2kπ-π≤12x-π3≤2kπ,k∈Z,解得4kπ-43π≤x≤4kπ+2π3,k∈Z,因为x∈[-2π,2π],所以函数的单调增区间为:(-43π,23π);故答案为:(-43π,23π).
∵y=f(x)的图象向右平移π3个单位长度后所得:y=cosω(x-π3)=cos(ωx-ωπ3);∵函数图象平移π3个单位长度后,所得的图象与原图象重合,说明函数平移整数个周期,就是2π的整数倍,所
把函数y=cos(x+4π3)的图象向右平移θ(θ>0)个单位,所得的函数为y=cos(x+4π3−θ),它是偶函数,所以θ=π3+kπ,k∈Z.故答案为:π3.
(1)∵cos2x=2cos^2x-1∴f(x)=1/2+cos(2x+π/6)/2对称轴2x0+π/6=π+2kπx0=5π/12+kπg(x0)=1+1/2sin(5π/6+2kπ)=5/4(2)
①f(x)=cos﹙2x-4π/3﹚+2cos²x=cos2xcos4π/3+sin2xsin4π/3+1+cos2x=1/2cos2x-√3/2sin2x+1=cos(2x+π/3)+1当
y'=3[cos(1/x)]^2*[cos(1/x)]'..=3[cos(1/x)]^2*[-sin(1/x)]*(1/x)'..=3[cos(1/x)]^2*[-sin(1/x)]*(-1/x^2)
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
X∈(0,3/π)x-π/6∈[-π/6,π/6]y=cos[2(x-π/6)]+2sin(x-π/6)=1-2sin^2(x-π/6)+2sin(x-π/6)令sin(x-π/6)=t(-1/2≤t