已知函数f(x)=sin^x+2√3sin(x+π/4)cos(x-π/4)-cos^2x-√3

求函数的最小正周期和单调递减区间

解：已知函数的表达式为：

f(x) = sin²x + 2√3sin(x+π/4)cos(x-π/4) -- cos²x -- √3 对吧？

细节分析：
① cos2x = cos(x + x)
= cosx cosx -- sinx sinx
= cos²x -- sin²x

② sin(x + π/4) = sinx cos(π/4) + cosx sin(π/4)
= (√2/2) sinx + (√2/2) cosx
= (√2/2) (sinx + cosx)

③ cos(x -- π/4) = cosx cos(π/4) + sinx sin(π/4)
= (√2/2) cosx + (√2/2) sinx
= (√2/2) (sinx + cosx)

把以上三式代入原函数表达式，得：
f(x) = sin²x + 2√3sin(x+π/4)cos(x-π/4) -- cos²x -- √3

= sin²x -- cos²x + 2√3sin(x+π/4)cos(x-π/4) -- √3

= -- (cos²x -- sin²x) + 2√3sin(x+π/4)cos(x-π/4) -- √3

= -- cos2x + 2√3 × [(√2/2) (sinx + cosx)]² -- √3

= -- cos2x + 2√3 × [ (1/2) × (sinx + cosx)² ] -- √3

= -- cos2x + √3 × (sinx + cosx)² -- √3

= -- cos2x + √3 × (1 + 2 sinx cosx) -- √3 (注：sin²x + cos²x = 1)

= -- cos2x + √3 × (1 + sin2x) -- √3 (注：sin2x = 2 sinx cosx)

= √3 sin2x -- cos2x

= 2 × [ (√3/2) sin2x -- (1/2) cos2x ]

= 2 × [ sin2x cos(π/6) -- cos2x sin(π/6) ]

= 2sin(2x -- π/6)

∴ 其最小正周期为：
T = 2π/ 2
= π

再求其单调递减区间：

设X = 2x -- π/6，

而sinX的单调递减区间为 [ 2kπ + π/2，2kπ + 3π/2 ]

∴ 2kπ + π/2 ≤ 2x -- π/6 ≤ 2kπ + 3π/2

∴2kπ + π/2 + π/6 ≤ 2x ≤ 2kπ + 3π/2 + π/6

∴2kπ + 2π/3 ≤ 2x ≤ 2kπ + 5π/3

∴ kπ + π/3 ≤ x ≤ kπ + 5π/6

∴ f（x）的单调递减区间为 [ kπ + π/3，kπ + 5π/6 ] （k ∈Z）。

祝您学习顺利！

1、 f(x)=-(cos²x-sin²x)+2√3cos[π/2-(x+π/4)]cos(x-π/4)-√3 =-cos2x+2√3cos(-x+π/4)cos(x-π/4)-√3 =-cos2x+2√3cos²(x-π/4)-√3 =-cos2x+2√3[1+cos2(x-π/4)]/2-√3 =-cos2x+√3[1+cos(2x-π/2)]-√3 =-cos2x+sin2x =√2(√2/2*sin2x-√2/2cos2x) =√2(sin2xcosπ/4-cos2xsinπ/4) =√2sin(2x-π/4) 所以t=2π/2=π 递减则2kπ+π/2<2x-π/4<2kπ+3π/2 2kπ+3π/4<2x<2kπ+7π/4 kπ+3π/8<x<kπ+7π/8 所以减区间(kπ+3π/8,kπ+7π/8) 2、 -π/12<=x<=25π/36 -5π/12<=2x-π/4<=41π/36 所以2x-π/4=-5π/12最小,x=-π/12 2x-π/4=π/2最大,x=3π/8 sin(-5π/12) =-sin(5π/12) =-sin(π/4+π/6) =-sinπ/4cosπ/6-cosπ/4sinπ/6 =-(√6+√2)/4 sinπ/2=1 所以 x=-π/12，y最小=-(√3+1)/2 x=3π/8，y最大=√2

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