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.................
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![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.................)
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