已知函数 f(x)=sinxcosx-√3sin^2x,①求 f(x)的最小正周期 ②求 f(x)在区间[0，π/2]上的最大值和最小值

\u5df2\u77e5\u51fd\u6570f(x)\uff1dsinxcosx\uff0d\u6839\u53f73sin\u7684\u5e73\u65b9x.\u6c42f(x)\u7684\u6700\u5c0f\u6b63\u5468\u671f\uff0c\u548c\u6c42f(x)\u5728\u533a

\u5316\u7b80\u516c\u5f0f
sinxcosx=1/2sin2x,
sin²x=1/2(1-cos2x)
\u8f85\u52a9\u89d2\u516c\u5f0f
f(x)=sinxcosx-\u221a3sin^2x,
=1/2sin2x-\u221a3/2(1-cos2x)
=1/2sin2x+\u221a3/2cos2x-\u221a3/2
=sin(2x+\u03c0/3)-\u221a3/2
f(x)\u7684\u6700\u5c0f\u6b63\u5468\u671fT=2\u03c0/2=\u03c0
\u6709\u5e2e\u52a9\u8bf7\u70b9\u597d\u8bc4\u6216\u8005\u91c7\u7eb3
\u795d\u4f60\u65b0\u7684\u4e00\u5b66\u671f\u5b66\u4e60\u8fdb\u6b65\uff01


\u2235x\u2208[0,\u03c0/2]
\u22342x\u2208[0,\u03c0]
2x+\u03c0/3\u2208[\u03c0/3,4\u03c0/3]
\u22342x+\u03c0/3=\u03c0/2\u65f6\uff0cf(x)max=1-\u221a3/2
2x+\u03c0/3=4\u03c0/3\u65f6\uff0cf(x)min=-\uff081+\u221a3\uff09/2

\u89e3: f(x) = -\u221a3sin²x + sinx cosx
= -\u221a3 \u00d7 (1 - cos(2x))/2 + (sin(2x))/2 (\u5229\u7528\u4e8c\u500d\u89d2\u516c\u5f0f)
= -\u221a3/2 + (\u221a3/2)cos(2x) + (1/2)sin(2x)
= -\u221a3/2 + sin(\u03c0/3) cos(2x) + cos(\u03c0/3) sin(2x)
= -\u221a3/2 + sin(\u03c0/3 + 2x) (\u5229\u7528\u4e24\u89d2\u548c\u7684\u6b63\u5f26\u516c\u5f0f)

(1) \u5468\u671f = 2\u03c0 / 2 = \u03c0 (T = 2\u03c0 / w)

（1）
化简公式
sinxcosx=1/2sin2x,
sin²x=1/2(1-cos2x)
辅助角公式
f(x)=sinxcosx-√3sin^2x,
=1/2sin2x-√3/2(1-cos2x)
=1/2sin2x+√3/2cos2x-√3/2
=sin(2x+π/3)-√3/2
f(x)的最小正周期T=2π/2=π
(2)
∵x∈[0,π/2]
∴2x∈[0,π]
2x+π/3∈[π/3,4π/3]
∴2x+π/3=π/2时，f(x)max=1-√3/2
2x+π/3=4π/3时，f(x)min=-（1+√3）/2

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