三角形练习
\u4e09\u89d2\u5f62\u7ec3\u4e60\u98981\u3001A\uff0cB\uff0cC\u4e3a\u4e09\u89d2\u5f62\u5185\u89d2\uff0c\u5df2\u77e51+cos2A-cos2B-cos2C=2sinBsinC\uff0c\u6c42\u89d2A
\u89e3\uff1a1+cos2A-cos2B-cos2C=2sinBsinC
2cos²A-1-2cos²B+1+2sin²C=2sinBsinC
cos²A-cos²B+sin² (A+B)=sinBsinC
cos²A-cos²B+sin²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC
cos²A-cos²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC
2cos²AsinB+2sinAcosAcosB=sin(180-A-B)
2cosA(cosAsinB+sinAcosB)-sin(A+B)=0
Sin(A+B)(2cosA-1)=0
cosA=1/2
A=60
2\u3001\u8bc1\u660e\uff1a(1+sin\u03b1+cos\u03b1+2sin\u03b1cos\u03b1)/(1+sin\u03b1+cos\u03b1)=sin\u03b1+cos\u03b1
1+sina+cosa+2sinacosa=sina+cosa+(sina+cosa)²
1+sina+cosa+2sinacosa=sina+cosa+1+2sinacosa
0=0\u6052\u6210\u7acb
\u4ee5\u4e0a\u5404\u6b65\u53ef\u9006\uff0c\u539f\u547d\u9898\u6210\u7acb
\u8bc1\u6bd5
3\u3001\u5728\u25b3ABC\u4e2d,sinB*sinC=cos²(A/2),\u5219\u25b3ABC\u7684\u5f62\u72b6\u662f?
sinBsin(180-A-B)=(1+cosA)/2
2sinBsin(A+B)=1+cosA
2sinB(sinAcosB+cosAsinB)=1+cosA
sin2BsinA+2cosAsin²B-cosA-1=0
sin2BsinA+cosA(2sin²B-1)=1
sin2BsinA-cosAcos2B=1
cos2BcosA-sin2BsinA=-1
cos(2B+A)=-1
\u56e0\u4e3aA\uff0cB\u662f\u4e09\u89d2\u5f62\u5185\u89d2
2B+A=180
\u56e0\u4e3aA+B+C=180
\u6240\u4ee5B=C
\u4e09\u89d2\u5f62ABC\u662f\u7b49\u8170\u4e09\u89d2\u5f62
4\u3001\u6c42\u51fd\u6570y=2-cos(x/3)\u7684\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\u5e76\u5206\u522b\u5199\u51fa\u4f7f\u8fd9\u4e2a\u51fd\u6570\u53d6\u5f97\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\u7684x\u7684\u96c6\u5408
-1\u2264cos(x/3)\u22641
-1\u2264-cos(x/3)\u22641
1\u22642-cos(x/3)\u22643
\u503c\u57df[1,3]
\u5f53cos(x/3)=1\u65f6\u5373x/3=2k\u03c0\u5373x=6k\u03c0\u65f6\uff0cy\u6709\u6700\u5c0f\u503c1\u6b64\u65f6{x\uff5cx=6k\u03c0,k\u2208Z}
\u5f53cos(x/3)=-1\u65f6\u5373x/3=2k\u03c0+\u03c0\u5373x=6k\u03c0+3\u03c0\u65f6\uff0cy\u6709\u6700\u5c0f\u503c1\u6b64\u65f6{x\uff5cx=6k\u03c0+3\u03c0,k\u2208Z}
5\u3001\u5df2\u77e5\u25b3ABC\uff0c\u82e5(2c-b)tanB=btanA\uff0c\u6c42\u89d2A
[(2c-b)/b]sinB/cosB=sinA/cosA
\u6b63\u5f26\u5b9a\u7406c/sinC=b/sinB=2R\u4ee3\u5165
(2sinC-sinB)cosA=sinAcosB
2sin(A+B)cosA=sinAcosB+cosAsinB
2sin(A+B)cosA-sin(A+B)=0
sin(A+B)(2cosA-1)=0
sin(A+B)\u22600
cosA=1/2
A=60\u5ea6
6\u3001\u5df2\u77e52cosx=3cosy\u6c42\u8bc1\uff1a3cosx-2cosy/2siny-3sinx=tan(x+y)
\u8bc1\u660e\uff1a3cosx-2cosy/2siny-3sinx=tan(x+y)
(3cosx-2cosy)/(2siny-3sinx)=sin(x+y)/cos(x+y)
(3cosx-2cosy)/(2siny-3sinx)=(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)
3cos²xcosy-3cosxsinxsiny-2cosxcos²y+2sinxcosxsiny=2sinxsinycosy+2sin²ycosx-3sin²xcosy-3sinxcosxsiny
3cos²xcosy+3sin²xcosy=2sin²ycosx+2cos²ycosx
3cosy(sin²x+cos²x)=2cosx(sin²y+cos²y)
3cosy=2cosx\u5df2\u77e5
\u6240\u4ee5\u4ee5\u4e0a\u5404\u6b65\u53ef\u9006
\u539f\u547d\u9898\u6210\u7acb
\u9700\u8981hi\u6211
\u5982\u679c\u8981\u4e09\u89d2\u5f62\u7ec3\u4e60\u9898\u8bdd\uff0c\u5efa\u8bae\u4f60\u53bb\u6587\u5e93\u53bb\u641c\u641c\uff0c\u6216\u8005\u5c06\u81ea\u5df1\u66fe\u7ecf\u505a\u8fc7\u7684\u9898\u62ff\u51fa\u6765\u7ffb\u7ffb\uff0c\u4e5f\u5f88\u6709\u6536\u83b7\u7684
可以利用多边形的内角和来求这道题具体解法：
解：CM‖FN
理由是：∵多边形ABCDEF为六边形
∴∠A+∠B+∠BCM+∠MCD+∠D+∠E+∠EFN+∠NFA=720°
又∵CM平分∠BCD,FN平分∠AFE
∴∠BCM=∠MCD,∠EFN=∠NFA 又∠A=∠D,∠B=∠C
∴2∠A+2∠B+2∠BCM+2∠EFN=720°
∴∠A+∠B+∠BCM+∠EFN=360°
又∵多边形ABCM为四边形
∴∠A+∠B+∠BCM+∠AMC=360°
∴∠EFN=∠AMC
∴CM‖FN
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