已知fx=sinx+cosx/sinxcosx且x属于0到二分之派求函数最小值
\u6c42\u51fd\u6570y=sinx-cosx+sinxcosx,x\u5c5e\u4e8e[0,\u03c0]\u7684\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\u8bbesinx-cosx=t,\u5219sinxcosx=(1-t*t)/2
0<=x<=\u03c0\uff0c\u5219-1<=t<=\u6839\u53f72
y=-t*t/2+t+1/2=-1/2(t-1)(t-1)+1
\u6700\u5c0f\u503c\u4e3a-1\uff0ct=-1,\u5373x=0
\u6700\u5927\u503c\u4e3a1\uff0ct=1,\u5373x=\u03c0/2\u6216\u03c0
\u89e3\uff1a(1)\u6362\u5143\u3002\u53ef\u8bbet=sinx-cosx=(\u221a2)sin[x-(\u03c0/4)].\u75310\u2264x\u2264\u03c0,\u53ef\u77e5-1\u2264t\u2264\u221a2,\u4e14t²=1-2sinxcosx.===>sinxcosx=(1-t²)/2.===>y=t+[(1-t²)/2]=1-(1/2)(t-1)².(-1\u2264t\u2264\u221a2).(2)y=1-0.5(t-1)².(-1\u2264t\u2264\u221a2).\u6613\u77e5\uff0cymin=y(-1)=-1,ymax=y(1)=1.
∵x∈（0，π/2）∴sinx＞0，cosx＞0
∴f(x)=(sinx+cosx)/(sinxcosx)＞0
令f(x)=(sinx+cosx)/(sinxcosx)=t＞0
t²=(sinx+cosx)²/(sinxcosx)²
= (sin²x+cos²x+2sinxcosx)/(sinxcosx)²
= (1+2sinxcosx)/(sinxcosx)²
= (1+sin2x)/(1/2sin2x)²
= (4+4sin2x)/(sin2x)²
t²(sin2x)²-4sin2x-4=0
sin2x=[4±√(4²+16t²)]/(2t²)
= [2±2√(1+t²)]/(t²)
x∈（0，π/2）
2x∈（0，π）
0＜sin2x≤1
0＜ [2±2√(1+t²)]/(t²)≤1
∵√(1+t²)＞1
∴2-2√(1+t²)＜0，舍去
∴ [2+2√(1+t²)]/(t²)≤1
∴2+2√(1+t²)≤t²
2√(1+t²)≤t²-2
4+4t²≤t^4-4t²+4
t^4-8t²≥0
t²≥8
t＞0
∴t≥2√2
即：f(x)=(sinx+cosx)/(sinxcosx)最小值2√2
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绛旓細褰搙鈭圼2k蟺,2k蟺+蟺/2)鈭(2k蟺+蟺/2,2k蟺+蟺]鏃讹紝鍥犱负sinx>0,f(x)=tgx,姝ゆ椂鍑芥暟鍦ㄦ瘡涓涓懆鏈熶笂閫掑 褰搙鈭圼2k蟺-蟺,2k蟺-蟺/2)鈭(2k蟺-蟺/2,2k蟺]鏃讹紝鍥犱负sinx<0,f(x)=-tgx,姝ゆ椂鍑芥暟鍦ㄦ瘡涓涓懆鏈熶笂閫掑噺
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