f(x)=1/2sin2xsinψ+COS²xcosψ-1/2sin(π/2+ψ),(0

f(x)=1/2sin2xsinψ+COS²xcosψ-1/2sin(π/2+ψ),(0
f(x)=1/2sin2xsinψ+COS²xcosψ-1/2sin(π/2+ψ),(0<ψ<π)其图像过点(π/6,1/2)求ψ

f(x)=1/2sin2xsinψ+COS²xcosψ-1/2sin(π/2+ψ),(0
f(π/6)=√3/4sinψ+3/4cosψ-1/2cosψ
=√3/4sinψ+1/4cosψ
=1/2(√3/2sinψ+1/2cosψ)
=1/2sin(ψ+π/6)
=1/2
sin(ψ+π/6)=1
ψ+π/6=2kπ+π/2
ψ=2kπ+π/3,k∈Z

cos2x=2cos²x-1, ∴cos²x=(1+cos2x)/2
sin(π/2+￡)=cos￡
∴f(x)=(1/2)sin2xsin￡+(1+cos2x)/2×cos￡-(1/2)cos￡
=(1/2)sin2xsin￡+(1/2)cos2xcos￡+(1/2)cos￡-(1/2)cos￡
=(1/2)(c...

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cos2x=2cos²x-1, ∴cos²x=(1+cos2x)/2
sin(π/2+￡)=cos￡
∴f(x)=(1/2)sin2xsin￡+(1+cos2x)/2×cos￡-(1/2)cos￡
=(1/2)sin2xsin￡+(1/2)cos2xcos￡+(1/2)cos￡-(1/2)cos￡
=(1/2)(cos2xcos￡+sin2xsin￡)=(1/2)cos(2x-￡)
x=π/6时,f(x)取得最大值, ∴cos(π/3-￡)=1
0<￡<π, -2π/3<π/3-￡<π/3, ∴π/3-￡=0
解得 ￡=π/3

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