问题: 求函数y=cos^2(x-pai/12) sin^2(x pai/12)-1周期．奇偶性
解答:
y=cos^2(x-π/12)+sin^2(x+π/12)-1
=cos^2(x-π/12)+sin^2(x+π/12)-[sin^2(x+π/12)+cos^2(x+π/12)]
=cos^2(x-π/12)-cos^2(x+π/12)
=[cos(x-π/12)-cos(x+π/12)][cos(x-π/12)+cos(x+π/12)]
=[sinxsin(π/12)][cosxcos(π/12)]
=(sinxcosx)[sin(π/12)cos(π/12)]
=(1/4)sin2xsin(π/6)
=(1/8)sin2x
所以：T=2π/2=π,是奇函数。
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