圆c的极坐标方程怎么化为普通方程 求详解 怎么把这个极坐标方程化为普通方程
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\u6240\u4ee5\u53ef\u4ee5\u63a8\u5bfc\uff1a
1\u3001\u03c1=4sin\u03b8\uff0c\u4e24\u8fb9\u540c\u4e58p\u53ef\u5f97
2\u3001\u3001\u03c1\u00d7\u03c1=4\u03c1sin\u03b8\uff0c\u516c\u793a\u4ee3\u6362\u53ef\u5f97
3\u3001x^2+y^2=4y
\u6781\u5750\u6807\uff1a
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\u6781\u5750\u6807\u65b9\u7a0b\uff1a
\u6781\u5750\u6807\u7cfb\u63cf\u8ff0\u7684\u66f2\u7ebf\u65b9\u7a0b\uff0c\u901a\u5e38\u8868\u793a\u4e3ar\u4e3a\u81ea\u53d8\u91cf\u03b8\u7684\u51fd\u6570\u3002\u6781\u5750\u6807\u65b9\u7a0b\u7ecf\u5e38\u4f1a\u8868\u73b0\u51fa\u4e0d\u540c\u7684\u5bf9\u79f0\u5f62\u5f0f\uff0c\u5982\u679cr(−\u03b8)=r(\u03b8)\uff0c\u5219\u66f2\u7ebf\u5173\u4e8e\u6781\u70b9(0\u00b0/180\u00b0)\u5bf9\u79f0\uff0c\u5982\u679cr(\u03c0+\u03b8)=r(\u03b8)\uff0c\u5219\u66f2\u7ebf\u5173\u4e8e\u6781\u70b9(90\u00b0/270\u00b0)\u5bf9\u79f0\uff0c\u5982\u679cr(\u03b8−\u03b1)=r(\u03b8)\uff0c\u5219\u66f2\u7ebf\u76f8\u5f53\u4e8e\u4ece\u6781\u70b9\u9006\u65f6\u9488\u65b9\u5411\u65cb\u8f6c\u03b1\u00b0\u3002
\u6781\u5750\u6807\u65b9\u7a0b\u63cf\u8ff0\u7684\u66f2\u7ebf\u65b9\u7a0b\u79f0\u4f5c\u6781\u5750\u6807\u65b9\u7a0b\uff0c\u901a\u5e38\u8868\u793a\u4e3ar\u4e3a\u81ea\u53d8\u91cf\u03b8\u7684\u51fd\u6570\u3002\u6781\u5750\u6807\u65b9\u7a0b\u7ecf\u5e38\u4f1a\u8868\u73b0\u51fa\u4e0d\u540c\u7684\u5bf9\u79f0\u5f62\u5f0f\uff0c\u5982\u679cr(−\u03b8) = r(\u03b8)\uff0c\u5219\u66f2\u7ebf\u5173\u4e8e\u6781\u70b9(0\u00b0/180\u00b0)\u5bf9\u79f0\uff0c\u5982\u679cr(\u03c0\uff0d\u03b8) = r(\u03b8)\uff0c\u5219\u66f2\u7ebf\u5173\u4e8e\u6781\u70b9(90\u00b0/270\u00b0)\u5bf9\u79f0\uff0c\u5982\u679cr(\u03b8−\u03b1) = r(\u03b8)\uff0c\u5219\u66f2\u7ebf\u76f8\u5f53\u4e8e\u4ece\u6781\u70b9\u9006\u65f6\u9488\u65b9\u5411\u65cb\u8f6c\u03b1\u00b0\u3002
\u5b9e\u9645\u4e0a\uff0c\u6781\u5750\u6807\u4e0e\u76f4\u89d2\u5750\u6807\u4e00\u6837\uff0c\u90fd\u662f\u4e3a\u4e86\u8868\u793a\u70b9\u5728\u7a7a\u95f4\u4e2d\u7684\u4f4d\u7f6e\u800c\u5f15\u5165\u7684\u53c2\u7167\u7cfb\u3002
\u6781\u5750\u6807\u662f\u7528\u8be5\u70b9\u5230\u5b9a\u70b9\uff08\u79f0\u4f5c\u6781\u70b9\uff09\u7684\u8ddd\u79bb\u53ca\u8be5\u70b9\u548c\u6781\u70b9\u7684\u8fde\u7ebf\u4e0e\u8fc7\u6781\u70b9\u7684\u5c04\u7ebf\uff08\u79f0\u4e3a\u6781\u8f74\uff09\u6240\u6210\u7684\u89d2\u5ea6\u6765\u786e\u5b9a\u5750\u6807\u7684\u3002
\u5173\u4e8e\u666e\u901a\u65b9\u7a0b\u4e0e\u6781\u5750\u6807\u65b9\u7a0b\u7684\u8f6c\u5316\uff0c\u53ea\u8981\u628a\u666e\u901a\u65b9\u7a0b\u7684x\u7528\u03c1cos\u03b8\u4ee3\u66ff\uff0c\u628ay\u7528\u03c1sin\u03b8 \u4ee3\u66ff\uff0c\u518d\u6574\u7406\uff0c\u5c31\u884c\u4e86\u3002
\u672c\u9898\u7684\u6781\u5750\u6807\u65b9\u7a0b\u548c\u666e\u901a\u65b9\u7a0b\u7684\u4e92\u5316\u5982\u4e0b\uff1a
\u03c1²=x²+y² x=\u03c1cos\u03b8
x²+y²=4x
x²-4x+y²=0
\uff08x-2\uff09²+y²=4
解：方程ρ=4sinθ可写成：ρ²=4ρsinθ，把y=ρsinθ，ρ²=x²+y²代入得：x²+y²=4y，
所以：x²+y²=4y即为所求；
方程ρcos(θ-π/4)=2√2可化为：ρ(cosθcosπ/4+sinθsinπ/4)=2√2，
化简得：ρcosθ+ρsinθ=4，把x=ρcosθ，y=ρsinθ代入得：x+y=4，
所以：x+y=4即为所求。
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