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答案:T=π 奇函数
y=1-(cos(x+π/12))^2+(cos(x-π/12))^2-1
=(cos(x-π/12))^2-(cos(x+π/12))^2
=(cos(x-π/12)-cos(x+π/12))×(cos(x-π/12)+cos(x+π/12))
=2cosxcosπ/12×2sinxsinπ/12
=(2sinxcosx)×(2sinπ/12cosπ/12)
=sin2x×sinπ/6
=(1/2)sin2x
所以最小正周期T=π
奇函数
y=1-(cos(x+π/12))^2+(cos(x-π/12))^2-1
=(cos(x-π/12))^2-(cos(x+π/12))^2
=(cos(x-π/12)-cos(x+π/12))×(cos(x-π/12)+cos(x+π/12))
=2cosxcosπ/12×2sinxsinπ/12
=(2sinxcosx)×(2sinπ/12cosπ/12)
=sin2x×sinπ/6
=(1/2)sin2x
所以最小正周期T=π
奇函数
1年前
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你能帮帮他们吗