(1)
因A+B+C=π
B+C=π-A
(B+C)/2=π/2-A/2
4sin²[(B+C)/2]-cos2A=3.5
可化为
4cos²A/2 - 2cos²A + 1 = 3.5
2(1+cosA) - 2cos²A = 2.5
cos²A - cosA + 0.25 = 0
则cosA=0.5
A=π/3
(2)
由正弦定理得
a/sinA=b/sinB=c/sinC
由比例的相关性质知道
a/sinA=(b+c)/(sinB+sinC)
√3/(√3/2)=3/(sinB+sinC)
sinB+sinC=3/2
而B+C=π-A=2π/3
sinB+sinC
=sinB+sin(2π/3-B)
=sinB + √3/2 cosB + 1/2 sinB
=√3(1/2 cosB + √3/2 sinB)
=√3sin(π/6+B)=3/2
因