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已知函数F(x)=sin(2x+π3)+sin(2x%π3)+2Cos2x...

f(x)=sin(2x+π/3)-√3cos(2x+π/3) =2sin2x. 因为sinx在2kπ-π/2≤x≤2kπ+π/2,k∈Z单增. 所以sin2x在2kπ-π/2≤2x≤2kπ+π/2,k∈Z单增. 解出x得递增区间(亲,请写成区间形式)。 因为sinx在2kπ+π/2≤x≤2kπ+3π/2,k∈Z单减. 所以sin2x在2kπ+π/2≤2x≤2kπ+3π/2...

(1)∵f(x)=sin2x?cos π 3 +cos2x?sin π 3 +sin2x?cos π 3 -cos2x?sin π 3 +cos2x=sin2x+cos2x= 2 sin(2x+ π 4 ),∴函数f(x)的最小正周期T= 2π 2 =π.(2)∵函数f(x)在区间[ - π 4 , π 8 ]上是增函数,在区间[ π 8 , π 4 ]上是减函数...

f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1 =2sin2xcosπ/3+2cos²x-1(和差化积) =sin2x+2cos²x-1 =sin2x+cos2x =√2*sin(2x+π/4) (1)f(x)的最小正周期T=2π/2=π (2)当x∈【-π/4,π/4】时,2x+π/4∈【-π/4,3π/4】 ∴ sin...

(1)∵f(x)=12cos2x+32sin2x-cos2x=32sin2x-12cos2x=sin(2x-π6),令2kπ≤2x-π6≤π2+2kπ,k∈Z,则kπ?π6≤x≤π3+kπ,k∈Z;∴函数f(x)的单调递增区间是[kπ?π6,π3+kπ],k∈Z;(2)列表如下;根据表中数据,建立平面直角坐标系,画出函数在一个周期...

已知函数f(x)=sin(2x+3\π)+sin(2x-3\π)+2cosx的平方-1求最小正周期和在(-4\π,4\π)上的最大值,最小值

解由f(x)=2cos(2x+2π/3)+√3sin2x =2cos2xcos(2π/3)-2sin2xsin2π/3+√3sin2x =2(-1/2)cos2x-2(√3/2)sin2x+√3sin2x =-cos2x 故函数的周期T=2π/2=π 函数的最大值为1.

2sin(x-π/4)sin(x+π/4) =2sin(x-π/4)cos[π/2-(x-π/4)] =2sin(x-π/4) cos(π/4-x) =2sin(x-π/4) cos(x-π/4) =sin(2x-π/2) =-sin(π/2-2x) =-cos2x ∴ f(x)=cos2xcosπ/3+sin2xsinπ/3-cos2x =-(cos2xcosπ/3-sin2xsinπ/3) =-cos(2x+π/3) T=2π/2=π 对称...

1.求函数fx的最大值和最小正周期 2.设A B C 为三角形ABC的三个内角若COSB=1/3 f(C/2)=-1/4 且C为锐角,求sinA f(x)=cos(2x+π/3)+sin² X=负二分之根号三sin2x+二分之一所以最大值为﹙√3 +1﹚/2最小正周期为π2.可知COSB=1/3 sinC=√3 /2 ...

解:①原式=f(x)=2cos2x+sinx^2 =2cos2x+1-cos2x/2 =3/2cos2x+1/2 故f(π/3)=3/2*cos2π/3+1/2 =-3/4+1/2 =-1/4 ②依f(x)=3/2cos2x+1/2可知 最大值为2 最小值为-1 很高兴为您解答,祝你学习进步!不懂可追问!

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