设函数f(x)=2sinxcosx-cos2x+1 求f(2/π) 求f(x)最大值和最小正周期
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解
f(x)=2sinxcosx-cos2x+1
=sin2x-cos2x+1
=√2(sin2xcosπ/4-cos2xsinπ/4)+1
=√2sin(2x-π/4)+1
∴
f(π/2)=√2sin(π/2×2-π/4)+1
=√2sin(3π/4)+1
=√2×(√2/2)+1
=1+1
=2
∵-1≤sin(2x-π/4)≤1
当sin(2x-π/4)=1时，f(x)取得最大值
∴f(x)max=√2×1+1=√2+1
∵T=2π/2=π
∴f(x)的最小正周期为：π
f(x)=2sinxcosx-cos2x+1
=sin2x-cos2x+1
=√2(sin2xcosπ/4-cos2xsinπ/4)+1
=√2sin(2x-π/4)+1
∴
f(π/2)=√2sin(π/2×2-π/4)+1
=√2sin(3π/4)+1
=√2×(√2/2)+1
=1+1
=2
∵-1≤sin(2x-π/4)≤1
当sin(2x-π/4)=1时，f(x)取得最大值
∴f(x)max=√2×1+1=√2+1
∵T=2π/2=π
∴f(x)的最小正周期为：π
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