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求y=sin^2(x+π/12)+cos^2(x-π/12)-1的最小正周期

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求y=sin^2(x+π/12)+cos^2(x-π/12)-1的最小正周期
求y=sin^2(x+π/12)+cos^2(x-π/12)-1的最小正周期
y=sin^2(x+π/12)+cos^2(x-π/12)-1
=[1-cos(2x+π/6)]/2+[1+cos(2x-π/6)]/2-1
=cos(2x-π/6)-cos(2x+π/6)
=(cos2xcosπ/6+sin2xsinπ/6)-(cos2xcosπ/6-sin2xsinπ/6)
=2sin2xsinπ/6
=2sin2x*(1/2)
=sin(2x)
sin函数的周期是2kπ,k=0,1,2...
所以sin2x的周期就是kπ,k=0,1,2,3...
要求的是正周期,只要正的部分,[kπ+0,kπ+π/2],k=0,1,2...