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!
