余弦定理证明

ABCDabc法一：作高法222222:sin,coscos,(sin)(cos)2cosAADBCBCDADbCCDbCBDBCCDabCABCcbCabCcababC解过点作交于点在直角三角形中由勾股定理得(bcosC,bsinC)(a,0)CxayOc22(cos)(sin0)bCabC22222cos2cossincbCabCabC2222coscababC法二:坐标法bc?AB解:以C为原点,BC为x轴建立直角坐标系ABCabc法三:向量法cab由三角形法则有22(cos)(sin0)bCabC,,CAbCBaABc令22||()ccab222222||22coscababcababC法四:正弦定理sinsinacAC　由得sinsin(2)BbC同理csinsin(1)cAaC()(2)BCAB利用代入消去角得22(1)A利用+(3)消去即得证coscos(3)cAbaC法五:正弦定理的推论222:2coscababC求证222:(2sin)(2sin)8sinsincosRARBRABC证明右边224sin()RAB右边2sincRC利用证明()CAB由得2222224(sincoscossin2sincossincos)cRABABAABB2222cos1sin,cos1sinAABB把代入得2222coscababC