作业帮在三角形ABC中,a,b,c分别是角A,B,C的对边,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) sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2) cos[(A-C)/2]怎么等于cos30
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![在三角形ABC中,a,b,c分别是角A,B,C的对边,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) sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2) cos[(A-C)/2]怎么等于cos30](/uploads/image/z/1601024-32-4.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ca%2Cb%2Cc%E5%88%86%E5%88%AB%E6%98%AF%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%2C%E8%AE%BEa%2Bc%3D2b%2CA-C%3D60%E5%BA%A6%2C%E6%B1%82sinB%E7%9A%84%E5%80%BC2sin%5B%28A%2BC%29%2F2%5D+%2A+cos%5B%28A-C%29%2F2%5D+%3D+2+%2A+2+%2A+sin%28B%2F2%29+%2A+cos%28B%2F2%29+sin%5B%28A%2BC%29%2F2%5D+%2A+cos30+%3D+2+%2A+sin%28B%2F2%29+%2A+cos%28B%2F2%29+cos%5B%28A-C%29%2F2%5D%E6%80%8E%E4%B9%88%E7%AD%89%E4%BA%8Ecos30)
在三角形ABC中,a,b,c分别是角A,B,C的对边,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) sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2) cos[(A-C)/2]怎么等于cos30
在三角形ABC中,a,b,c分别是角A,B,C的对边,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)
sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)
cos[(A-C)/2]怎么等于cos30
在三角形ABC中,a,b,c分别是角A,B,C的对边,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) sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2) cos[(A-C)/2]怎么等于cos30
1、根据正弦定理 a/sinA＝b/sinB＝c/sinC
得：a＝（sinA/sinB）*b c＝（sinC/sinB)*b
将其带入已知条件 a+c＝2b中
可得sinA+sinC＝2sinB
根据三角函数和公式
sinA+sinC＝2sin[(A+C)/2] * cos[(A-C)/2]
∴A+B+C=∏
∵sin[(A+C)/2]＝sin[(∏－B)/2]＝sin（∏/2－B/2)＝cos（B/2)
∴A-C＝60°
∵cos[(A-C)/2]＝cos30°＝（√3）/2
∵sinA+sinC＝√3*cos（B/2)＝2sinB
根据倍角公式 sinB＝2sin（B/2)cos（B/2)
√3*cos（B/2)＝4sin（B/2)cos（B/2)
sin（B/2)＝（√3）/4
cos（B/2)＝√（1-(（√3）/4)^2)
=（√13）/4
sinB＝2sin（B/2)cos（B/2)=（√39）/8
A-C=60度 你自己列出来了啊