函数y=cos²(x- π/12) + sin²(x+ π/12)-1 的最小正周期是?

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函数y=cos²(x- π/12) + sin²(x+ π/12)-1 的最小正周期是?
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函数y=cos²(x- π/12) + sin²(x+ π/12)-1 的最小正周期是?
函数y=cos²(x- π/12) + sin²(x+ π/12)-1 的最小正周期是?

函数y=cos²(x- π/12) + sin²(x+ π/12)-1 的最小正周期是?
y=cos²(x-π/12) + sin²(x+π/12)-1
=(1+cos(2x-π/6))/2+(1-cos(2x+π/6))/2-1
= cos(2x-π/6)/2-cos(2x+π/6)/2
=1/2[cos(2x-π/6)- cos(2x+π/6)]
=1/2[cos2x cosπ/6+sin2x sinπ/6-( cos2x cosπ/6-sin2x sinπ/6)]
=1/2[2sin2x sinπ/6]
= sin2x sinπ/6
=1/2 sin2x
∴函数的最小正周期是π.