已知函数f(x)= sin2x-2sinx^2 ,求f(x)的最小正周期
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(Ⅰ)因为f(x)=sin2x-(1-cos2x)=2sin(2x+π4)-1,所以函数f(x)的最小正周期为T=2π2=π(Ⅱ)由(Ⅰ)知,当2x+π4=2kπ−π2,即x=kπ−π8(k∈Z)时,
f(x)=sin2x-cos2x+1=√2*(√2/2*sin2x-√2/2*cos2x)+1=√2sin(2x-π/4)+1最小正周期为:T=2π/2=π∵-1≤sin(2x-π/4)≤1∴1-√2
A=2,T=π∴ω=2∴f(x)=2sin(2x+φ﹚过﹙π/6,2﹚∴2sin(π/3+φ﹚=2sin(π/3+φ﹚=1π/3+φ=2kπ+π/2φ=2kπ+π/6∴φ=π/6∴f(x)=2sin(
f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,
1.sinx≠0,∴x≠kπ.∴f(x)的定义域为{x|x≠kπ,k∈Z}2.f(x)=(sin2x-cos2x+1)/(2sinx)=(2sixcosx+2sin²x)/(2sinx)=c
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解.(1)函数f(x)=2sin2x+sin2x-1=sin2x-cos2x=2sin(2x-π4)…(3分)所以f(x)的最小正周期是π,…(4分)当2x-π4=2kπ+π2,k∈Z,即x=kπ+3
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