sinB sinc=2cos(A B)

a/sinA=b/sinB=c/sinC=2RS△ABC=(ab/2)·sinC=(bc/2)·sinA=(ac/2)·sinB=abc/(4R)故S=(ab/2)·sinC=1/2a*asinB/s

由正弦定理得sinB=b*(sinA/a)sinC=c*(sinA/a)代入得(1/2)*a^2*[(sinBsinC)/sinA]=(1/2)*a^2*[(sinA*bc)/a^2]=(1/2)*b

三角形面积公式为：S=(1/2)abSinC=(1/2)acSinB=(1/2)bcSinA证：已知S=(1/2)a²sinBsinC/sinA由正弦定理：a/SinA=b/SinB=c/S

令k=a/sinA=b/sinBb=ksinB因为S=1/2absinC=1/2a*ksinBsinC=1/2a*(a/sinA)sinBsinC=1/2*a^2*sinBsinC/sinA

S=1/2*absinC这个公式吧,他是由bsinA是高乘以底a得来的现在只要证出1/2*absinC=1/2*a^2*(sinBsinC)/(sinA)就可以了也就是bsinC=a*(sinBsin

a/sinA=b/sinB=>b=a*sinB/sinAS=1/2absinC=1/2a*a*sinBsinC/sinA=1/2a^2sinBsinC/sinA

^2+c^2-a^2=2bccosA=bc,cosA=1/2,A=60°(sinB)^2+(sinC)^2-3/4=sinBsinC=3/4(sinB)^2+(sinC)^2=3/2(sinB)^2+

设三角形的顶点为A、B、C,对应的边长为a、b、c.过顶点B做AC边上的垂线,设垂线长度为h,则有h=asinC.SΔABC＝h*b/2=absinC/2正弦定理a/sinA=b/sinB可得b=as

cos²(A/2)=(1+cosA)/2=sinBsinC1+cos(180-B-C)=2sinBsinC1-cos(B+C)=2sinBsinC1-(cosBcosC-sinBsinC)=

等边三角形由sin²A=sinBsinC得a²=bc(*)把2a=b+c变形为a=(b+c)/2代入(*)式,得（b-c)²=0∴b=c,再代入(*)式得a=b∴三角形A

cos(B+C)=cosBcosC-sinBsinC=-2分之1,∴B+C=120°∴A=60°

a/sinA=b/sinB=c/sinC=t(at)^2=(bt)^2+(bt*ct)+(ct)^2a^2=b^2+bc+c^2b^2+c^2-a^2=-bccosA=(b^2+c^2-a^2)/(2

由三角形ABC中sinA：a=sinB:b=sinC:c得a^2=b^2+bc+c^21/2=(b^2+c^2-a^2)/2bc又由cosA=(b^2+c^2-a^2)/2bccosA=1/2A=60

2sinBsinC－cos(B－C)=2sinBsinC－(cosBcosC+sinBsinC)=sinBsinC－cosBcosC=-cos(B+C)=1/2cos(B+C)=-1/20

由a/sinA=b/sinB=c/sinC所以sin²A=sinBsinC则a²=bc2a=b+c则a²=(b+c)²/4=bcb²+c²+

B=π/3ksinBsinC=cos²A-cos²B+sin²C降次：2ksinBsinC=cos2A-cos2C+1-cos2B√3ksinC=2sin(A+C)sin

倍角公式sin2x=2sinxcosx所以左边=[2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]=sin(A+B)*sin(A-B)

稍等再答：你的题目是不是搞错了？再答：Sin=SinBSinC?再问：好像是，不好意思再问：sin²A再问：平方咋个打不起？再问：sin²A再问：真的打不起平方再问：再答：应该是S

∵sinBsinC=cos2A/2∴1＋cosA＝2sinBsinC∴2sinBsinC－cosA＝1,即2sinBsinC＋cos(B＋C)＝1,即cos(B－C)＝1∵在△ABC中,－π＜B－C＜