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函数y=sin(π/3+x)的最小正周期为

2pai除以2等于pai

y=sinx+sin(x-π/3) =sinx+1/2sinx-√3/2cosx =3/2sinx-√3/2cosx =√3(√3/2sinx-1/2cosx) =√3sin(x-π/60 最小正周期是:T=2π 最大值是是:√3

∵y=1-sin 2 ( x+ π 3 )= 1 2 + 1 2 cos(2x+ 2π 3 )∴T= 2π 2 =π故答案为:π

就这样

y=3sin(π/2-x)+4sin(2π+x) =3cosx+4cosx =7cosx T=2π/1=2π

因为:y=sin(ωx+π/4)的最小正周期是2π/3 所以:2π/ω=2π/3 解得:ω=3

sin(π/3+4x)=cos[π/2-(π/3+4x)]=cos(π/6-4x) f(x)=sin(π/3+4x)+cos(4x-π/6)=2cos(4x-π/6) 最小正周期是:T=2π/4=π/2 2kπ=

由于函数y=3sinx+sin(x+π2)=3sinx+cosx=2(32sinx+12cosx)=2sin(x+π6),故函数的最小正周期是 2π,故答案为 2π.

Y=5sin(xπ/3 -1/4) =5sin(πx/3 -1/4) 故T=2π/(π/3)=6 故最小正周期为6.

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