正弦定理、余弦定理复习课任后兵定理正弦定理余弦定理内容a2=b2=c2=b2+c2-2bccosAc2+a2-2cacosBa2+b2-2abcosC1.正、余弦定理RCcBbAa2sinsinsin变形(1)a=,b=,c=;(2)sinA=,sinB=,sinC=;(3)a∶b∶c=.cosA=;cosB=;cosC=2RsinB2RsinCsinA∶sinB∶sinCb2+c2-a22bcc2+a2-b22aca2+b2-c22aba2Rc2Rb2R2RsinA2.三角形的面积公式3.常用结论BacAbcCabSABCsin21sin21sin21(1)符号:(2)有关与角AAAAtan,cos,0sinBA(3)CBACBACBAtan)tan(,cos)cos(,sin)sin((4)(5)222acb2sin)2cos(,2cos)2sin(CBACBABABAbacoscossinsin)2,0(A),2(A222;acb①在△ABC中,,则②在△ABC中,则最大内角的余弦值为(1)三角形中边角判断4:2:3sin:sin:sinCBA41()()22sin53cosBA,434或B22sin54sinBA4:2:3::cba(2)三角形解的个数的判断①在△ABC中,若,②在△ABC中,若,则此三角形有两解.4,32,60caA则B等于603,1,30baA()()23sinB1sinC(3)三角形形状的判断①(教材第10页)在△ABC中,,BbAacoscos则此三角形是等腰三角形.②在△ABC中,若③在△ABC中,若,BABAcoscossinsin则此三角形是钝角三角形.222acb则此三角形是锐角三角形.()()()BBAAcossincossin0)cos(BA典型例题讲解例1.(教材第20页第14题改编)△ABC的内角A,B,C的对边分别为a,b,c,且cAbBa)coscos(3求证:(I)222)(3cba(II)BAtan2tan解:(I)角化边得:cbcacbbacbcaa223222222(II)边化角得:CABBAsin)cossincos(sin3)sin(BA整理得:BABAsincos2cossinBBAAcossin2cossin0cos0cosBA(II)BAtan2tan例2.(2014·陕西卷改编)已知a,b,c为△ABC的内角A,B,C的对边,若a,b,c成等差数列,且ACsin2sin求cosB的值.解:accab22ab23161122)23()2(cos222aaaaaB在△ABC中,若变式:a,b,c成等比数列,且23cos)cos(BCA,求B.解:由题意得:acb223)cos()cos(CACA23sinsin2CACABsinsinsin223sin22B3BB角大小在中间!例3.(教材第20页中线公式改编)6,5,4BCACAB,D是BC上一点,且BD=2DC,求AD.在△ABC中,法二:利用4445ADCADBcoscos法一:先在△ABC求cosB,再在△ABD中求AD.14ADABCD2例4.已知a,b,c为△ABC的内角A,B,C的对边,若bCaCasin33cos(I)求A;(II)若a=2,△ABC的面积为,求b,c..3.解:(I)边化角得:BCACAsinsinsin33cossin)sin(CA整理得:CACAsincossinsin333tanA0sinC(II)由及面积公式得:AbcSsin21433sin21bcbc由余弦定理有:Abccbacos22223cos2422bccbbcbccb2)(24cb2cb求得,(II)若a=2,△ABC的面积为,求b,c.33A变式:已知a,b,c为△ABC的内角A,B,C的对边,若cbCaCasin3cos.(I)求A;(II)若,求b+c.3,1aABCA.解:(I)边化角得:CBCACAsinsinsinsin3cossinCCAsin)sin(整理得:CCACAsinsincossinsin30sinC1cossin3AA1)6sin(2A66AA0(II)若,求b+c.3,1aABCA(II)由有,即:1ABCA1)cos(AABCA21cosbcAbc由余弦定理有:3cos2322bccbbccb3)(23cb知己知彼,百战不殆不入虎穴,焉得虎子滴水穿石,绳锯木断螳螂捕蝉,黄雀在后怎样解题认真读题执行解题计划需要耐心回顾小结