三角形ABC,cosC (cosA-根号三sinA)cosB

sinA=5/13＜1/2,故可以是A＞150°,或A＜30°1/2＜cosB＜√2/2,故45°＜B＜60°由于A+B＜180°所以A为锐角故cosA=12/13,sinB=4/5cosC=-cos

cosB=1/2B=60sinB=根号3/2sinA=2/3cos^2A=(1-(2/3)^2)cosA=根号5/3orCosA=-根号5/3cosC=cos(180-A-B)=-cos(A+B)=s

根据正弦定理a/sinA=b/sinB=c/sinC,原式可变形为:bsin2A+bcos2A=跟号3a,即b=根号3a.将上面的结果带入余弦定理“cosC=(a^2+b^2-c^2)/(2·a·b)

∵m//n∴﹙√3b-c﹚×cosA－a×cosC＝0,根据余弦定理得,﹙√3b-c﹚×﹙b²＋c²－a²﹚／2bc＝a×﹙a²＋b²－c²

已知三角形ABC的三个内角,满足A+B=2B,设x=cos(A-C)/2,f(x)=cosB(1/cosA+1/cosC)已知三角形ABC的三个内角,满足A+B=2B,设x=cos(A-C)/2,f(

ABC依次成等差数列=>2B=A+C=>3B=180°=>B=60°=>A+C=120°1/cosA+1/cosC=-√2/cosB=>(cosA+cosC)/cosAcosC=√2cos(A+C)带

利用积化和差和和差化积B=60,A+C=1201/cosA+1/cosC=-2√2(cosA+cosC)/(cosAcosC)=-2√2cosA+cosC=2cos[(A+C)/2]cos[(A-C)

因为:A+B+C=180,A+C=2B,3B=180,B=60度1/cosA+1/cosC=-√2/cosB,(cosC+cosA)/cosA*cosC=-√2/cosB,利用公式:cosA+cosC

三角形ABC中,A+C=2B,得：B=60,A+C=1201/cosA+1/cosC=-√2/cosB=-2√2cosA+cosC=-2√2cosAcosC（左边用和差化积,右边用积化和差）2cos[

题目叙述有错误,想必应该是：“cosB=根号10除以10”吧?!在此基础上,解答过程如下：∵cosB=根号10除以10∴sinB=3·根号10除以10∴cosC=-cos(A+B)=sin45°·si

A+B=π-Ccos(A+B)=cos(π-C)=-cosCcos[(A+B)/2]=cos[(π-C)/2]=cos(π/2-C/2)=sin(C/2)sin(3A+3B)=sin3*(A+B)=s

∵a＞b,∴A＞B.作∠BAD＝B交边BC于点D.设BD＝x,则AD＝x,DC＝5-x.在ΔADC中,注意cos∠DAC＝cos(A-B)=31/32,由余弦定理得：(5-x)^2=x^2+4^2-2

2sinA(cosC/2)^2+2sinC（cos(A/2)）^2=3sinBsinA(1+cosC)+sinC(1+cosA)=sinA+sinC+sin(A+C)=sinA+sinC+sinB=3

,{sin(A-B)+sinC)/{cos(A-B)+cosC}=,{sin(A-B)+sin(A+B))/{cos(A-B)-cos(A+B)}=2sinAcosB/2sinAsinB=cosB/s

2种方法如下：、①∵a＞b,∴A＞B.作∠BAD＝B交边BC于点D.设BD＝x,则AD＝x,DC＝5-x.在ΔADC中,注意cos∠DAC＝cos(A-B)=31/32,由余弦定理得：(5-x)^2=

等于

∵a＞b,∴A＞B.作∠BAD＝B交边BC于点D.设BD＝x,则AD＝x,DC＝5-x.在ΔADC中,注意cos∠DAC＝cos(A-B)=31/32,由余弦定理得：(5-x)^2=x^2+4^2-2

∵a＞b,∴A＞B.作∠BAD＝B交边BC于点D.设BD＝x,则AD＝x,DC＝5-x.在ΔADC中,注意cos∠DAC＝cos(A-B)=31/32,由余弦定理得：(5-x)^2=x^2+4^2-2

根据正弦定理：a/sinA=b/sinB,∵a=5,b=4∴设sinA=x,sinB=4x/5∵sin²A+cos²A=1∴cos²A=1-x²,cos

根据正弦定理,(a+b)/a=sinB/(sinB-sinA)=(sinA+sinB)/sinA∴sinA·sinB=(sinB+sinA)(sinB-sinA)=2sin[(B+A)/2]·cos[