三角函数题(要简单步骤)已知函数f(x)=√3sin(2x-π/6)+2sin^2(x-π/12)求函数的最小正周期求使

1.
2sin^2(x-π/12) = 1 - cos[2·(x-π/12)] = 1 - cos(2x-π/6) ,∴f(x) = 2·{(√3/2)sin(2x-π/6) - (1/2)·cos(2x-π/6) + 1/2} = 2sin[(2x-π/6) - (π/6)] + 1 = 1 + 2sin(2x - π/3) ,最小正周期 = 2π/2 = π ,函数最大值为3 ,此时 sin(2x - π/3) = 1 ,2x - π/3 = 2kπ + π/2 ,x = kπ + 5π/12 ,k∈Z .
2.
A、B∈(0 ,π) ,∴sinA、sinB > 0 ,根据余弦值代入同角关系式得：sinA = 12/13 ,sinB = 4/5 ,∴sinC = sin(A+B) = sinAcosB + cosAsinB = 16/65 ,由正弦定理 ,BC/sinA = AB/sinC ,解得AB = 4/3 ,∴S(△ABC) = (1/2)·AB·BC·sinB = 8/3 .
3.
结合已知等式并根据余弦定理：a^2 = b^2 + c^2 - 2bc·cosA = b^2 + c^2 - √3bc ,∴cosA = √3/2 ,∴A = π/6 .2sinBcosC-sin(B-C) = 2sinBcosC - sinBcosC + cosBsinC = sinBcosC + cosBsinC = sin(B+C) = sinA = 1/2 .

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