在三角形ABC中，若2B=A+C则sinA*sinC=cosB*cosB，三角形ABC的面积=4*根号下3，求三边a，b，c
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u82e52B=A+C\u5219sinA*sinC=cosB*cosB\uff0c\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef=4*\u6839\u53f7\u4e0b3\uff0c\u6c42\u4e09\u8fb9a\uff0cb\uff0cc\u2235A+B+C=180\u00b0\uff0c2B=A+C\uff0c
\u2234B=60\u00b0
sinAsinC=cos²B
sinAsinC=1/4
sinAsin\uff08A+\u03c0/3\uff09=1/4
1/2sin²A+\u6839\u53f73/2sinAcosA=1/4
1/4\uff081-cos2A\uff09+\u6839\u53f73/4sin2A=1/4
\uff081-cos2A\uff09+\u6839\u53f73sin2A=1
\u6839\u53f73sin2A-cos2A=0
2sin\uff082A-\u03c0/6\uff09=0
sin\uff082A-\u03c0/6\uff09=0
\u22350<A<120\u00b0
\u2234-30\u00b0\uff1c2A-30\u00b0\uff1c210\u00b0
\u22342A-\u03c0/6=0\u00b0\u6216180\u00b0
\u2234A=15\u00b0\uff0cC=105\u00b0\u6216A=105\u00b0\uff0cC=15\u00b0
S\u25b3ABC=1/2acsinB=4\u6839\u53f73
\u2234ac=16
c=asinC/sinA=\uff082+\u6839\u53f73\uff09a\u6216\uff082-\u6839\u53f73\uff09a
\u4ee3\u4ebaac=16
\u89e3\u5f97a=\uff0cc=
cosB=a²+c²-b²/2ac=1/2
\u89e3\u5f97b=
2B=A+C
A+B+C=3B=180\u00b0\uff0cB=60\u00b0
sinA*sinC=1/4
sin[180-(B+C)]*sinC=1/4
sin(B+C)*sinC=1/4
\u63a5\u4e0b\u6765\u5c31\u662f\u5316\u7b80
1/2sin(2C-\u03c0/6)=0
C=\u03c0/12
A=7\u03c0/12
S=1/2acsinB=4\u6839\u53f73
ac=16
a/sinA=b/sinB=c/sinC
a=bsinA/sinB,c=bsinC/sinB
b^2*sinAsinC/sin^2(B)=16
b^2cos^2(B)/sin^2=16
b=4\u6839\u53f73
a=bsinA/sinB=2\u6839\u53f76+2\u6839\u53f72
c=bsinC/sinB=2\u6839\u53f76-2\u6839\u53f72
B=60 A+C=120
SINA*SINC=1/2*[COS(A-C)-COS(A+C)]=1/2
得：COS(A-C)=1/2 A-C=60
A=90 C=30
所以：a=2c=根号3/2b
S=bc/2=4根号3
得：a=4根号2 b=2根号6 c=2根号2
2B=A+C A+B+C=180
B=60 A+C=120
sinA*sinC=1/2*[cos(A-C)-cos(A+C)]=cosB*cosB=1/4
cos(A-C)=0
A-C=90
A=105, C=15
又S=(1/2)acsinB=（√3）／4ac（√3为根号3）
=4√3
ac=16
再由a/sinA=b/sinB=c/sinC
可求得答案（貌似小烦了，呵呵）这个三角形出得实在有点怪
（PS:一楼的，sinA*sinC=1/2,cosB*cosB=1/4,明显有问题呃）
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