三角形ABC中，a、b、c、分别是角A、B、C的对边，设a+b=2b，A-C=三分之派，求sinB的值
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0ca\uff0cb\uff0cc\u5206\u522b\u662f\u89d2A\uff0cB\uff0cC\u7684\u5bf9\u8fb9\uff0c\u8bbea\uff0bc\uff1d2b\uff0cA\uff0dC\uff1d\u4e09\u5206\u4e4b\u03c0\uff0c\u6c42SinB\u7684\u503c\u3002\u89e3a\uff0bc\uff1d2b\u7531\u6b63\u5f26\u5b9a\u7406\u5f97sinA\uff0bsinC\uff1d2sinB
\u53732sinB=sinA\uff0bsinC
\u6709\u79ef\u5316\u548c\u5dee\u516c\u5f0f\u5f972sinB=sinA\uff0bsinC=2sin\uff08A+C)/2cos\uff08A-C)/2
\u7531A+B+C=\u03c0\uff0c\u53732sinB=2sin\uff08\u03c0-B)/2cos\uff08\u03c0/3)/2
\u5373sinB=sin\uff08\u03c0-B)/2cos\uff08\u03c0/3)/2
sinB=cosB/2cos\uff08\u03c0/6)
\u53732sinB/2cosB/2=cosB/2cos\uff08\u03c0/6)
sinB/2=\u221a3/4
cosB=1-2(sinB/2)²=1-2*\uff08\u221a3/4)²=10/16=5/8
sinB=\u221a\uff5b\u30101-(cosB)²\u3011=\u221a39/8
\u7531\u6b63\u5f26\u5b9a\u7406\uff1aa/sinA=b/sinB=c/sinC\uff0c
a+c=2b\uff0c
\u2014\u2014\u300bsinA+sinC=2sinB\uff0c
A+C=\u03c0-B\uff0cA-C=\u03c0/3\uff0c
\u2014\u2014\u300bA=2\u03c0/3-B/2\uff0cC=\u03c0/3-B/2\uff0c
\u2014\u2014\u300b2sinB=sinA+sinC
=sin(2\u03c0/3-B/2)+sin(\u03c0/3-B/2)
=v3/2*cosB/2+1/2*sinB/2+v3/2*cosB/2-1/2*sinB/2
=v3*cosB/2\uff0c
\u2014\u2014\u300b2*2sinB/2*cosB/2=v3*cosB/2\uff0c
\u2014\u2014\u300bsinB/2=v3/4\uff0c
\u2014\u2014\u300bcosB/2=v13/4\uff0c
\u2014\u2014\u300bsinB=2sinB/2*cosB/2=v39/8\u3002
您的题目有误。
原题：
在三角形ABC中，a，b，c分别是角A，B，C的对边，设a＋c＝2b，A－C＝三分之π，求SinB的值。
解答：
解：
a＋c＝2b由正弦定理得sinA＋sinC＝2sinB
即2sinB=sinA＋sinC
有积化和差公式得2sinB=sinA＋sinC=2sin（A+C)/2cos（A-C)/2
由A+B+C=π，即2sinB=2sin（π-B)/2cos（π/3)/2
即sinB=sin（π-B)/2cos（π/3)/2
sinB=cosB/2cos（π/6)
即2sinB/2cosB/2=cosB/2cos（π/6)
sinB/2=√3/4
cosB=1-2(sinB/2)²=1-2*（√3/4)²=10/16=5/8
sinB=√｛【1-(cosB)²】=√39/8
祝楼主学习进步
你的题有问题没？？a+b=2b??,怎么不写成a=b?有区别么，这种题我看别问了，最基本的三角函数题，这都不会的话，以后就危险了，多做两个想想，会出来的。。加油啊
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