作业帮咋解这道三角函数在三角形ABC中,设a＋c＝2b,A－C＝60度,求SinB的值?2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2) 这个我看不懂.................
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 06:57:50
![咋解这道三角函数在三角形ABC中,设a＋c＝2b,A－C＝60度,求SinB的值?2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2) 这个我看不懂.................](/uploads/image/z/6998360-32-0.jpg?t=%E5%92%8B%E8%A7%A3%E8%BF%99%E9%81%93%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%AE%BEa%EF%BC%8Bc%EF%BC%9D2b%2CA%EF%BC%8DC%EF%BC%9D60%E5%BA%A6%2C%E6%B1%82SinB%E7%9A%84%E5%80%BC%3F2sin%5B%28A%2BC%29%2F2%5D+%C3%97+cos%5B%28A-C%29%2F2%5D+%3D+2+%C3%97+2+%C3%97sin%28B%2F2%29+%C3%97cos%28B%2F2%29+%E8%BF%99%E4%B8%AA%E6%88%91%E7%9C%8B%E4%B8%8D%E6%87%82.................)
咋解这道三角函数在三角形ABC中,设a＋c＝2b,A－C＝60度,求SinB的值?2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2) 这个我看不懂.................
咋解这道三角函数
在三角形ABC中,设a＋c＝2b,A－C＝60度,求SinB的值?
2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2) 这个我看不懂.................
咋解这道三角函数在三角形ABC中,设a＋c＝2b,A－C＝60度,求SinB的值?2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2) 这个我看不懂.................
∵在三角形ABC中,
∴a/sinA = b/sinB = c/sinC
∵a + c = 2b
∴sinA + sinC = 2sinB
∵ sinA + sinC =sin[(A+C)/2+(A-C)/2]+sin[(A+C)/2-(A-C)/2]
=2sin[(A+C)/2] × cos[(A-C)/2]
sinB =2 ×sin(B/2) ×cos(B/2)
sinA + sinC = 2sinB
∴2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2)
sin[(A+C)/2] × cos30 = 2 × sin(B/2) ×cos(B/2)
cos[90 - (A+C)/2] × cos30 = 2 ×sin(B/2)× cos(B/2)
cos(B/2) × cos30 = 2×sin(B/2) ×cos(B/2)
cos30 = 2 ×sin(B/2)
∴sin(B/2) = √3 / 4
cos(B/2)=√13/4
∴sinB=2×√3/4×√13/4=√39/8
我已经补充了回答