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已知函数y=sin(12x+π3),x∈[?2π,2π](1)求最小...

(1)y=sin(12x+π3)最小正周期T=2π12=4π…..(4分)(2)由-π2+2kπ≤12x+π3≤π2+2kπ,k∈Z得:-5π3+4kπ≤x≤π3+4kπ,k∈Z…(10分)∵x∈[-2π,2π],∴所求函数的单调递增区间为[-5π3,π3]…(12分)

(1)令z=12x+π3,函数y=sinz的单调递增区间是[?π2+2kπ,π2+2kπ],k∈Z,由?π2+2kπ≤12x+π3≤π2+2kπ,得?5π3+4kπ≤x≤π3+4kπ,k∈Z,设A=[-2π,2π],B={x|?5π3+4kπ≤x≤π3+4kπ,k∈Z},可得A∩B=[?5π3,π3],∴f(x)的单调递增区间为[?5π3,π3];(2)...

(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...

(1)∵函数y=2sin(12x+π3)且 T=2π12=4π对称轴方程满足:12x+π3=π2+kπ,k∈Z即对称轴方程为:x=π3+2kπ,k∈Z∵对称中心的横坐标满足:12x+π3=kπ,k∈Z则x=-2π3+2kπ,k∈Z即对称中心的坐标是(-2π3+2kπ,0)k∈Z;(2)由2kπ-π2≤12x+π3≤2kπ+π2(k∈Z)得...

(1)函数y=12sin(2x+π6)的振幅为12,周期为π,初相为π6.(2)列表: x ?π12 π6 5π12 2π3 11π12 2x+π6 0 π2 π 3π2 2π y=12sin(2x+π6) 0 12 0 ?12 0画简图:(3)函数y=sinx的图象向左平移 π6个单位,得到函数y=sin(x+π6)的图象,再保持纵坐...

(1)令2x+π3=kπ+π2可得x=kπ2+π12,k∈Z,∴函数y=f(x)的图象的对称轴为x=kπ2+π12,k∈Z,取k=0可得t的最小值为π12;(2)当x0∈[-π12,π6]时,2x0+π3∈[π6,2π3],∴sin(2x0+π3)∈[12,1],∴f(x0)∈[2,3]要使mf(x0)-2=0成立,只需f(x0)=2m,...

(1)∵f(x)=2sin(x?π12),∴f(?π6)=2sin(?π6?π12)=2sin(?π4)=?2sin(π4)=-1.(2)∵sinθ=?45,θ∈(3π2,2π),∴cosθ=1?sin2θ=35,∴sin2θ=2sinθcosθ=?2425,∴cos2θ=2cos2θ?1=?725,∴f(2θ+π3)=2sin(2θ+π4

y=sin2x+cos2(x-π3)=1?cos2x+cos(2x?2π3)+12=12cos(2x-2π3)-12cos2x+1=12(-12cos2x+32sin2x-cos2x)+1=12(32sin2x-32cos2x)+1=32sin(2x-π3)+1,当2kπ-π2≤2x-π3≤2kπ+π2(k∈Z)时,即kπ-π12≤x≤kπ+5π12(k∈Z)时,函数单调增.∴增区间为[k...

y=3sinxcosx+sin2x-12=32sin2x-12cos2x=sin(2x-π6)(1)函数f(x)的最小正周期:2π2=π.由?π2+2kπ≤2x?π6≤2kπ+π2,k∈Z,解得?π6+kπ≤x≤kπ+π3 k∈Z,函数单调递增区间[?π6+kπ,kπ+π3],k∈Z.(2)将f(x)的函数图象纵坐标不变,横坐标变为原来...

(1)y=sin2x+2sinxsin(π2?x)+3sin2(3π2?x)=sin2x+2sinxcosx+3cos2x=sin2x+2sinxcosx+3cos2xsin2x+cos2x=tan2x+2tanx+3tan2x+1∵tanx=12∴y=14+1+314+1=175(2)由(1)y=sin2x+2sinxsin(π2?x)+3sin2(3π2?x)=sin2x+2sinxcosx+3cos2x=2+sin2x+co...

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