已知函数fx=2cosx(sinx+ccosx) (1)求f(5π/4)值
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f（x）=2sinx（sinx+cosx）
=2sin²x+2sinxcosx
=-(1-2sin²x)+2sinxcosx+1
=-cos2x+sin2x+1
=√2[(√2/2)*sin2x-(√2/2)cos2x]+1
=√2[(sin2xcos(π/4)-cos2xsin(π/4)]+1
=√2sin(2x-π/4)+1
f（5π/4）=
√2sin(9π/4)+1
=
2
函数f（x）的最小正周期
t=2π/2=π
单调递增区间
[-√2+1,√2+1]
=2sin²x+2sinxcosx
=-(1-2sin²x)+2sinxcosx+1
=-cos2x+sin2x+1
=√2[(√2/2)*sin2x-(√2/2)cos2x]+1
=√2[(sin2xcos(π/4)-cos2xsin(π/4)]+1
=√2sin(2x-π/4)+1
f（5π/4）=
√2sin(9π/4)+1
=
2
函数f（x）的最小正周期
t=2π/2=π
单调递增区间
[-√2+1,√2+1]
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