Question

已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos（B+C）】-cos2A=7/2

Answer 1

先解出cosA

因为cos(B+C) = cos(π -A) = - cosA

cos2A = 2cos²A -1

解方程

2(1+cosA) -(2cos²A-1) = 7/2

化简即　2cos²A - 2cosA + 1/2 =0

cosA = 1/2

(1)

2sinBcosC = sin(B+C) - sin(C-B) = sinA - sin(C-B)

所以　 sinA - sin(C-B) = sinA

sin(C-B) =0

C=B

又因为A=π/3 , 所以C=B= (π-π/3)/2=π/3

ABC是等边三角形

(2)

余弦定理

a² = b² +c² -2bc cosA = b²+c² -bc

所以　3 = b²+c² -bc

b+c=3, 所以(b+c)² = b²+c² +2bc =9

两式相减

3bc = 9-3 =6

bc =2

故bc =2,

b+c =3

解之得

b=1, c=2或

b=2, c=1