在三角形ABC中，角A.B.C的对边分别为a.b.c.已知a+b=51c=根号7。且4sin平方2分之A+B减去cos2C=2分之7（1)求
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a+b=5
c=√7
4sin^2[(A+B)/2] - cos2C=7/2
4sin^2(90°-C/2) - (2cos^2C-1)=7/2
4cos^2(C/2) - (2cos^2C-1) = 7/2
2(cosC+1) - (2cos^2C-1) = 7/2
2cosC+2 - 2cos^2C +1 = 7/2
cos^2C-cosC +1/4 = 0
(cosc-1/2) = 0
cosC = 1/2
C = 60°
c^2=a^2+b^2-2abcosC = a^2+b^2-ab = (a+b)^2-3ab
ab = [ (a+b)^2-c^2 ] / 3 = (5^2-7)/3 = 6
S = 1/2absinC = 1/2*6*sin60° = 3√3/2
c=√7
4sin^2[(A+B)/2] - cos2C=7/2
4sin^2(90°-C/2) - (2cos^2C-1)=7/2
4cos^2(C/2) - (2cos^2C-1) = 7/2
2(cosC+1) - (2cos^2C-1) = 7/2
2cosC+2 - 2cos^2C +1 = 7/2
cos^2C-cosC +1/4 = 0
(cosc-1/2) = 0
cosC = 1/2
C = 60°
c^2=a^2+b^2-2abcosC = a^2+b^2-ab = (a+b)^2-3ab
ab = [ (a+b)^2-c^2 ] / 3 = (5^2-7)/3 = 6
S = 1/2absinC = 1/2*6*sin60° = 3√3/2
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