sin^2 (x+π/12)+sin^2 (x-π/12)

sin^2 (x+π/12)-sin^2 (x-π/12)
学过二倍角公式吧！
cos2A=1-2sin^2 (A)
利用此公式,
sin^2 (x+π/12)-sin^2 (x-π/12)
=[1-cos(2x+π/6)]/2-[1-cos(2x-π/6)]/2
=[cos(2x-π/6)-cos(2x+π/6)]/2
=[(cos2xcosπ/6+sin2xsinπ/6)-(cos2xcosπ/6-sin2xsinπ/6)]/2
=[根号(3)sin2x]/2

y∈[1,3]
当y=1时，sin(x+π/3)=-1，x+π/3=2kπ-π/2，x=kπ-5π/12，k∈Z
当y=3时，sin(x+π/3)=1， x+π/3=2kπ+π/2，x=kπ+π/12，k∈Z
∴当x∈[kπ-5π/12,kπ+π/12]时，y∈[1,3]

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