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已知函数F(x)=sin(2x+π6),x∈R.(1)求函数F...

(1)T=2π2=π.(2)由2kπ-π2≤2x+π6≤2kπ+π2,得kπ-π3≤x≤kπ+π6,k∈Z,∴函数的单调增区间为[kπ-π3,kπ+π6](k∈Z).(3)∵x∈[0,π2],∴2x+π6∈[π6,7π6],∴-12≤sin(2x+π6)≤1,∴当2x+π6=π2,即x=π6时函数有最大值1,当2x+π6=7π6时,即x=π2,函数有...

参考

(1)函数f(x)=sin(2x+π6)+sin(2x-π6)+2cos2x=sin2xcosπ6+cos2xsinπ6+sin2xcosπ6-cos2xsinπ6+1+cos2x=sin2x+cos2x+1=2sin(2x+π4)+1.由2kπ+π2≤2x+π4≤2kπ+3π2,解得kπ+π8≤x≤kπ+5π8(k∈Z).∴函数f(x)的单调递减区间[kπ+π8,kπ+5π8](k∈Z...

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

1. f(x)=sin(2x+π/6)+1/2 则最小正周期为T=2π/2=π。 2. y=(1/2)(cosx)^2+(√3/2)sinxcosx+1 =(√3/4)sin2x+(1/4)cos2x+5/4 =(1/2)[(√3/2)sin2x+(1/2)cos2x]+5/4 =(1/2)(sin2xcosπ/6+cos2xsinπ/6)+5/4 =(1/2)sin(2x+π/6)+5/4 f(x)最小正周期为T=2π/...

(1)根据函数 f(x)=2sin(2x- π 6 ),x∈R ,可得函数的最小正周期为 2π 2 =π,f(0)=2sin(- π 6 )=2×(- 1 2 )=-1.(2)令 2kπ- π 2 ≤2x- π 6 ≤2kπ+ π 2 ,k∈z,求得 kπ- π 3 ≤x≤kπ+ π 3 ,故函数的增区间为[kπ- π 3 ,kπ+ π 3 ],k∈z.(...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

(1)令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,可得 kπ-π3≤x≤kπ+π6,k∈z,故函数的增区间为:[kπ?π3,kπ+π6],k∈Z.(2)当x∈[0,π2]时,π6≤2x+π6≤7π6,-12≤sin(2x+π6)≤1,故f(x)的最大值为2+a+1=4,解得a=1.

因为这里书写不便,故将我的答案做成图像贴于下方,谨供楼主参考(若图像显示过小,点击图片可放大)

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