正弦定理和余弦定理在△ABC中,设a+c=2b.A-C=π/3,求sinB的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 05:10:17
正弦定理和余弦定理在△ABC中,设a+c=2b.A-C=π/3,求sinB的值
正弦定理和余弦定理
在△ABC中,设a+c=2b.A-C=π/3,求sinB的值
正弦定理和余弦定理在△ABC中,设a+c=2b.A-C=π/3,求sinB的值
因为 a + c = 2b
由正弦定理,知:
sinA +sinC = 2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB
sin[(A+C)/2] * cos(π/6) = sinB
因为A + B + C = 180
所以:(A+C)/2 = π/2 - B/2
所以:
cos(B/2) * √3/2 = 2sin(B/2)cos(B/2)
显然B/2不等于π/2,cos(B/2)不等于0
所以:
sin(B/2) = √3/4
cos(B/2) = √13/4
sinB = 2sin(B/2)cos(B/2) = √39/8
a+c=2b得,sinA+sinC=2sinB
2sin(A+C/2)cos(A-C/2)=4sinB/2cosB/2
因为A-C=π/3,(A+C)/2+B/2=π/2
化简下得 根号3cosB/2=4sinB/2cosB/2
sinB/2=根号3/4 cosB/2= 根号13/4
sinB=2sinB/2cosB/2=根号39/8
a+c=2b,C=π/3 - B/2 ,A=2π/3 - B/2
∴ sinA+sinC=2sinB
(√3/2)cosB/2 +(1/2)sinB/2 +(√3/2)cosB/2 - (1/2)sinB/2
=4sinB/2 .cosB/2
√3=4sinB/2
sinB/2=√3/4 cosB/2=√13/4
sinB=2 (√3/4)(√13/4)=√39/8