如何用向量的方法证明正弦和余弦定理？ 如何用向量的方法证明正弦和余弦定理？
\u7528\u5411\u91cf\u7684\u65b9\u6cd5\u8bc1\u660e\u6b63\u5f26\u5b9a\u7406\u6b65\u9aa41
\u8bb0\u5411\u91cfi \uff0c\u4f7fi\u5782\u76f4\u4e8eAC\u4e8eC\uff0c\u25b3ABC\u4e09\u8fb9AB,BC,CA\u4e3a\u5411\u91cfa,b,c
\u2234a+b+c=0
\u5219i\uff08a+b+c\uff09
=i\u00b7a+i\u00b7b+i\u00b7c
=a\u00b7cos(180-(C-90))+b\u00b70+c\u00b7cos(90-A)
=-asinC+csinA=0
\u63a5\u7740\u5f97\u5230\u6b63\u5f26\u5b9a\u7406
\u5176\u4ed6
\u6b65\u9aa42.
\u5728\u9510\u89d2\u25b3ABC\u4e2d\uff0c\u8bbeBC=a,AC=b,AB=c\u3002\u4f5cCH\u22a5AB\u5782\u8db3\u4e3a\u70b9H
CH=a\u00b7sinB
CH=b\u00b7sinA
\u2234a\u00b7sinB=b\u00b7sinA
\u5f97\u5230a/sinA=b/sinB
\u540c\u7406\uff0c\u5728\u25b3ABC\u4e2d\uff0c
b/sinB=c/sinC
\u6b65\u9aa43.
\u8bc1\u660ea/sinA=b/sinB=c/sinC=2R\uff1a
\u4efb\u610f\u4e09\u89d2\u5f62ABC,\u4f5cABC\u7684\u5916\u63a5\u5706O.
\u4f5c\u76f4\u5f84BD\u4ea4\u2299O\u4e8eD. \u8fde\u63a5DA.
\u56e0\u4e3a\u76f4\u5f84\u6240\u5bf9\u7684\u5706\u5468\u89d2\u662f\u76f4\u89d2,\u6240\u4ee5\u2220DAB=90\u5ea6
\u56e0\u4e3a\u540c\u5f27\u6240\u5bf9\u7684\u5706\u5468\u89d2\u76f8\u7b49,\u6240\u4ee5\u2220D\u7b49\u4e8e\u2220C.
\u6240\u4ee5c/sinC\uff1dc/sinD=BD=2R
\u7c7b\u4f3c\u53ef\u8bc1\u5176\u4f59\u4e24\u4e2a\u7b49\u5f0f\u3002
\u25b3ABC\u4e3a\u9510\u89d2\u4e09\u89d2\u5f62\uff0c\u8fc7\u70b9A\u4f5c\u5355\u4f4d\u5411\u91cfj\u5782\u76f4\u4e8e\u5411\u91cfAC\uff0c\u5219j\u4e0e\u5411\u91cfAB\u7684\u5939\u89d2\u4e3a90\u00b0-A\uff0cj\u4e0e\u5411\u91cfCB\u7684\u5939\u89d2\u4e3a90\u00b0-C
\u2235AC+CB=AB
\u5728\u5411\u91cf\u7b49\u5f0f\u4e24\u8fb9\u540c\u4e58\u5411\u91cfj\uff0c\u5f97\uff1a
j\u00b7\uff08AC+CB\uff09=j\u00b7AB
\u2234\u2502j\u2502\u2502AC\u2502cos90\u00b0+\u2502j\u2502\u2502CB\u2502cos(90\u00b0-C)
=\u2502j\u2502\u2502AB\u2502cos(90\u00b0-A)
\u2234asinC=csinA \uff08AB\u7684\u6a21=c\uff0ccos\uff0890º-C\uff09=sinC\uff09\uff08CB\u7684\u6a21=a\uff0ccos\uff0890º-A\uff09=sinA
\u2234a/sinA=c/sinC
\u540c\u7406\uff0c\u8fc7\u70b9C\u4f5c\u4e0e\u5411\u91cfCB\u5782\u76f4\u7684\u5355\u4f4d\u5411\u91cfj\uff0c\u53ef\u5f97
c/sinC=b/sinB
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\u5df2\u77e5\u25b3ABC\uff0c\u8bbeBC=a\uff0cCA=b\uff0cAB=c\uff0c\u4f5cAD\u22a5BC\uff0c\u5782\u8db3\u4e3aD.
\u5219Rt\u25b3ADB\u4e2d\uff0csinB=AD/AB
\u2234AD=AB\u00b7sinB=csinB
\u2234S\u25b3ABC=1/2a\u00b7AD=1/2acsinB
\u540c\u7406\u53ef\u8bc1S\u25b3ABC=1/2absinC=1/2bcsinA
\u5373S\u25b3ABC=1/2acsinB=1/2absinC=1/2bcsinA
\u2234absinC=bcsinA=acsinB
\u5728\u7b49\u5f0f\u4e24\u7aef\u540c\u65f6\u9664\u4ee5ABC\uff0c\u53ef\u5f97sinC/c=sinB/b=sinA/a
△ABC为锐角三角形，过点A作单位向量j垂直于向量AC，则j与向量AB的夹角为90°-A，j与向量CB的夹角为90°-C
∵AC+CB=AB
在向量等式两边同乘向量j，得：
j·（AC+CB）=j·AB
∴│j││AC│cos90°+│j││CB│cos(90°-C)
=│j││AB│cos(90°-A)
∴asinC=csinA （AB的模=c，cos（90º-C）=sinC）（CB的模=a，cos（90º-A）=sinA
∴a/sinA=c/sinC
同理，过点C作与向量CB垂直的单位向量j，可得
c/sinC=b/sinB
扩展资料
利用三角形的面积证明余弦定理
已知△ABC，设BC=a，CA=b，AB=c，作AD⊥BC，垂足为D.
则Rt△ADB中，sinB=AD/AB
∴AD=AB·sinB=csinB
∴S△ABC=1/2a·AD=1/2acsinB
同理可证S△ABC=1/2absinC=1/2bcsinA
即S△ABC=1/2acsinB=1/2absinC=1/2bcsinA
∴absinC=bcsinA=acsinB
在等式两端同时除以ABC，可得sinC/c=sinB/b=sinA/a
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