咋解这道三角函数
问题描述：
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.................
答
∵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