cos(a c)等于cosb

答：b²=accos(A-C)+cosB+cos²B=cos(A-C)-cos(A+C)+cos²(A+C)=cosAcosC+sinAsinC-cosAcosC+sin

=0sina+sinb=-sinycosa+cosb=-cosysinasinb=0.5((sina+sinb)^2-1)=0.5(siny^2-1)cosacosb=0.5(cosy^2-1)cos

cosA·cosB=1所以cosA=cosB=1或者cosA=cosB=-1否则cosAcosB

sina+sinb=1/2cosa-cosb=1/3分别平分得(sina+sinb)^2=sin^2a+2sinasinb+sin^2b=1/41(cosa-cosb)^2=cos^2a-2cosac

sinα+sinβ=1/2①cosα+cosβ=1/3②①²+②²:sin²α+2sinαsinβ+sin²β+cos²α+2cosαcosβ+cos

直角三角形中,角ACB等于九十度,BC等于a,AC等于b,AB等于c,则sinA等于cosB等于tanA等于sinA=a/ccosB=a/ctanA=a/

由b^2=ac知道a,b.c成等比数列,则SinASinBSinC也成等比数列Cos(A-C)+CosB=Cos(A-C)-Cos(A+C)展开得2SinASinC=3/2得出2倍SinB的平方=3/

经推算,此题只能求出外接圆的半径.

cos(A-C)cosB=2/3积化和差0.5[cos(A-C+B)+cos(A-C-B)]=2/30.5[cos(π-2C)+cos(2A-π)]=2/3-0.5[cos2C+cos2A]=2/3再

^2=ac,所以a,b.c成等比数列,则SinASinBSinC也成等比数列Cos(A-C)+CosB=Cos(A-C)-Cos(A+C)展开得2SinASinC=3/2得出2倍SinB的平方=3/2

cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2sinAsinC=3/4根据正弦定理,

5/3再问：为什么不等于负的。。。不是有两个值吗？再答：三角形里角的度数是小于180度的，则SINB是大于0的并且cosB=五分之三可以得出角B是0＜B＜90°的再问：为什么0＜B＜90再答：三角形里

应该是cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2]吧.

cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2sinAsinC=3/4根据正弦定理,

sina+sinb=1/2(sina+sinb)^2=1/4sina^2+2sinasinb+sinb^2=1/4(1)cosa+cosb=1/3(cosa+cosb)^2=1/9cosa^2+2co

cos(A-C)+cosB=cos(A-C)-cos(A+C)=3/22sinAsinC=3/2sinAsinC=3/4b^2=ac,所以(sinB)^2=sinAsinC=3/4sinB=根号3/2

cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2即sinAsinC=3/4根据正弦定理

条件给了cos(A-C)+cosB=3/2,后面算出B的两种可能120或60,假设B=120,cosB=-1/2,将其代入原式cos(A-C)+cosB=cos(A-C)-1/2=3/2即：cos(A

cos[(7/5)b]和cosb之间没有倍数关系啊