在△ABC中,a、b、c分别为角A、B、C所对的边,a=2√3,tan[(A+B)/2]+tan(C/2)=4,sinBsinC=cos^2(A/2),求A、B及b、c
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在△ABC中,a、b、c分别为角A、B、C所对的边,a=2√3,tan[(A+B)/2]+tan(C/2)=4,sinBsinC=cos^2(A/2),求A、B及b、c
在△ABC中,a、b、c分别为角A、B、C所对的边,a=2√3,tan[(A+B)/2]+tan(C/2)=4,sinBsinC=cos^2(A/2),求A、B及b、c
在△ABC中,a、b、c分别为角A、B、C所对的边,a=2√3,tan[(A+B)/2]+tan(C/2)=4,sinBsinC=cos^2(A/2),求A、B及b、c
tan[(A+B)/2] + tan(C/2) = 4
tan[(π-C)/2] + tan(C/2) =
cot[C/2] + tan(C/2) = 4
[cos(C/2)]^2 + [sin(C/2)]^2 = 4sin(C/2)cos(C/2)
sin C = 1/2
sinB sinC = cos^2[(π-B-C)/2]
(sinB)/2 = sin^2 [(B+C)/2]
sinB = 2 sin^2 [(B+C)/2] = 1 - cos(B+C)
sinB = 1 - cosBcosC + sinBsinC
(sinB)/2 = 1 - cosBcosC
[(sinB)/2 -1]^2 = (cosBcosC)^2
(sinB)^2 /4 - sinB + 1 = [1-(sinB)^2]×(3/4)
4(sinB)^2 - 4sinB + 1 = 0
( 2sinB - 1)^2 = 0
解得:sin B = 1/2
如果 C=150°,那么 B=30°,A=0° 不成立
所以 B = C = 30° A = 120°
a = 2根号3
b = c = 2