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(1)求函数y=sin(12x+π6)的最小正周期与单调递增...

(1)f(x)=2sin(ωx-π6)sin(ωx+π3)=2sin(ωx-π6)sin[(ωx-π6)+π2]=2sin(ωx-π6)cos(ωx-π6)=sin(2ωx-π3),∵T=π,∴ω=1,∴f(x)=sin(2x-π3),∵x∈[π8,5π12],∴2x-π3∈[-π12,π2],根据正弦函数在此区间单调递增,得到:f(x)min=sin...

(1) f(x)=sin(2(x-π/12)) x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6] 单调递减区间是2(x-π/12)∈[π/2,3π/2] 即x-π/12∈[π/4,3π/4] 则x∈[π/3,5π/6] (2)x∈[-π/12,π/2]时, 2(x-π/12)∈[-π/3,5π/6] 而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),...

有一道基本相同 已知函数f(x)=2sin(ωx),其中常数ω>0(1)若y=f(x)在[-π4,2π3]上单调递增,求ω的取值范围;(2)令ω=2,将函数y=f(x)的图象向左平移π6个单位,再向上平移1个单位,得到函数y=g(x)的图象,区间[a,b](a,b∈R,且a<b...

(1)∵f(x)=cos +2sin ·sin = cos2x+ sin2x+(sinx-cosx)(sinx+cosx)= cos2x+ sin2x+sin 2 x-cos 2 x= cos2x+ sin2x-cos2x=sin .∴周期T= = . 由 =k + (k∈Z),得x= (k∈Z).∴函数图象的对称轴方程为x= (k∈Z).(2)∵x∈ ,∴ ∈ .∵f(x)=sin 在区间 ...

函数的周期T=2πω=2π2=π,由-π2+2kπ≤2x+π3≤π2+2kπ,解得?5π12+kπ≤x≤π12+kπ,即函数的递增区间为[?5π12+kπ,π12+kπ],k∈Z,由2x+π3=π2+2kπ,即x=π12+kπ,k∈Z,即函数的对称轴为x=π12+kπ,k∈Z,由2x+π3=kπ,即x=-π6+kπ2,即函数的对称中心为(-π6+...

(1)f(x)=1+cos(2ωx?π3)2?1?cos2ωx2=12[cos(2ωx?π3)+cos2ωx]=12[(12cos2ωx+32sin2ωx)+cos2ωx]=12(32sin2ωx+32cos2ωx)=32(12sin2ωx+32cos2ωx)=32sin(2ωx+π3).由题意可知,f(x)的最小正周期T=π,∴2π|2ω|=π,又∵ω>0,∴ω=1,∴f(x)=32sin...

(I)令u=2x- π 6 ,则函数y=3sinu的单调增区间为[- π 2 +2kπ, π 2 +2kπ]k∈Z(5分)由- π 2 +2kπ≤2x- π 6 ≤ π 2 +2kπ,得:- π 6 +kπ≤x≤ π 3 +kπk∈Z函数y=3sin(2x- π 6 )的单调增区间为:[- π 6 +kπ, π 3 +kπ]k∈Z(8分)(II)∵x∈ [- π 12 ...

(Ⅰ)因为f(x)=(2cos2x?1)sin2x+12cos4x=12sin4x+12cos4x=22sin(4x+π4)∴T=2π4=π2,函数的最大值为:22.(Ⅱ)∵f(x)=22sin(4x+π4),f(α)=22,所以sin(4α+π4)=1,∴4α+π4=π2+2kπ,k∈Z,∴α=π16+kπ2,又∵α∈(π2, π),∴α=916π.

(1)∵f(x)=2sin(ωx),当ω=12时,F(x)=f(x)+f(x+π)=2sinx2+2sin(x2+π2)=2sinx2+2cosx2=22sin(x2+π4)由?π2+2kπ≤x2+π4≤π2+2kπ,k∈Z得:x∈[?3π2+4kπ,π2+4kπ],(k∈Z),即得f(x)的递增单调区间为:[?3π2+4kπ,π2+4kπ],(k∈Z),由π2+2kπ≤x2...

(1)令z=12x+π3,则y=sinz,y=sinz的单调递减区间为[2kπ+π2,2kπ+3π2],k∈Z,由2kπ+π2≤12x+π3≤2kπ+3π2,k∈Z,得:4kπ+π3≤x≤4kπ+7π3,k∈Z,又z=12x+π3在R上为增函数,故原函数的单调递减区间为:[4kπ+π3,4kπ+7π3]k∈Z,(2)令z=12x+π3,则y=sin...

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