余弦定理与正弦定理的等价性.docx
余弦定理与正弦定理的等价性甘肃省合水县第一中学薛树英邮编：745400 信箱：xshy2008126余弦定理与正弦定理是解斜三角形的依据，二者是互通的，是等价的.由余弦定理推得正弦定理中，由 cZm/+a?-IcibcOSCr得a2-b2-c2cos C =2absinC=Vl-cos2C =+b)2V 2abJ(a+b+c)(a+b-c)(a-b+c)(-a+b+c)2ab2js(s-a)(s-b)(s-c),其中 _ a + + c ab2cabc _ sinC 2/s(s-a)(s-b)(s-c)同理，有b _abcsinB 2js(s-a)(s-b)(s-c)a _abcsinA 2s(s-a)(s-b)(s-c)abc sinA sinB sinC由正弦定理推得余弦定理AABC中，设其外接圆半径为R,有a b c=2RsinA sinB sinCa=2RsinA , b=2RsinB , c=2RsinCa2 +b2 -2abcosC=(2Rsin A)2 +(2RsinB)2-2 2RsinA 2RsinB - cosC=4R2 (sin2 A+ sin2B- 2 sinAsinBcosC)=4R2fsin2A+ sin2B+ 2sinAsinBcos(A + B)4 R2 (sin2 A+ sin2B+ ；sin 2 Asin 2 B-2sin 2Asin2B)c2 = (2 RsinC) = 4R2sin2(A+ B) = 4R2(sinAcosB+cosAsinB)2=4R2(sin2Acos2B+ cos2Asin2B+sin 24sin 2B)= 4R2sin2A(l-sin2B) + (l-sin2A)sin2B+sin 2 Asin 2B)2=4 R2(sin2A+ sin2B+ ； sin 2 Asin 2B-2 sin2Asin2B)例题在,C中，/八5。,心向COSC-亚c2 =q2 +Z72 -2ahcosC（I）求5c边的长;（ID若点I）是AB的中点，求线段CD的长度.（2006年全国高考数学H文科17题）解法1 ： ( I )由 26，得 J5 cosC= sinC=55sinA=sin18() -(45 +C)=sin(45 +C)= (cosC + sinC)2也（2也也屈由正弦定理知,T - io. a回义巫fiC= AOsinA = 10_ = 3sinB V2（II）由正弦定理，得小阻 = 2sinB V22也可以根据由余弦定理得，/IB2 =BC2 + /IC2-2BCMCcosC2 石=18 +10-2x372 xVl()x- = 4AB = 2-BD = 1 AB = 1 .2由余弦定理知， CD2 =BD2+BC2-2 BD- BC, cos B= l + 18-2xlx3V2x = 132CD = 713-解法2: ( I )由正弦定理知,AV10 XAB=ACsinC=5 =2sinB V2T由2加，倚石.cosC= sinC=55cosA=cosI80 -(45 +C)=-cos(45 +C)=- (cosC + sinC)V2一骂.巫510. BC2 = AB2 + AC2- 2A B- AC- cos Ar JlO= 4 + 10-2x2x710x(-) = 1810(II)求AB时，应用余弦定理.BC2 = AB2+AC2-2ABACcosA/. 18 = AB2+10 + 2AB即 AB2+2AB-8 = 0AB = 2,或 AB = -4(舍)BD = j AB = 12然后,应用余弦定理求得“cD = g解法3: ( I )同解法2.(II)求AB时，仍然应用余弦定理.AC2 =AB2+BC2-2AB- BC-cos B10=AB2+18 + 2AB. AB = 2,或 AB = 4即 AB26AB+8 = 0当 AB = 2时,BI) 1AB21,根据余弦定理求得，CD = V13 -当AB = 4时,在AABC中,c°sB = R = 3，c°sC = 逅 =与22V25 V5cosB<cosC,所以,B>C, AOAB,而 AC = /0，AB=4 显然不成立.
