在三角形ABC中,a、b、c分别是角A、B、C的对边,且1+cos（π+2A）=2sin2[﹙B+C﹚÷2]

(1)1+cos(π+2A)=1-cos2A=2sin^2A=2(1-cos^2A)=2sin^2[(B+C)/2]=1-cos(B+C)=1+cosA
=>cosA=1/2 =>A=60°
(2)由正弦定理,a/sinA=b/sinB=c/sinC=4√3
=>b=4√3sinB,c=4√3sinC
=>S=1/2*b*c*sinA=12√3sinBsinC
sinBsinC=-1/2[cos(B+C)－cos(B-C)]=1/2[cos(B-C)－1/2]
当cos(B-C)=1,即B=C=60°时,sinBsinC取最大值,此时面积即有最大值S=3√3.
ΔABC为等边三角形

1+cos（π+2A）=2sin2[﹙B+C﹚÷2] 变形：
1-cos2A = 2 sin2[﹙π-A﹚÷2]
1-(1-2*SinA*SinA) = 2 sin﹙π-A﹚
2* SinA*SinA = 2 sin﹙A﹚
Sin(A)=1,
A= π/2 {90度}
面积Sabc= b * c /2 = aSinB * aCosB / 2 = (...

(1)1+cosπcos2A-sinπsin2A=2sin2[B/2+C/2]
1+cosπcos2A-sinπsin2A=2sin(B+C)
1-cos2A=2sinA
2sin平方A=2sinA
sinA=1
A=90°
（2）cosA=(c平方+b平方-a平方)/(2cb)
c平方+b平方-a平方=0
c平方+b平方=36

等边三角形