已知△ABC和三内角A、B、C成等差数列,A为最小角,且√3cosA/2=sinA+sinC,求A、B、C的值.A,B,C成等差数列 ==> A+C=2B而 A+B+C =180°∴ A+B+C = 3B=180 ==> B=60°√3cos(A/2) = sinA+sinC==>√3cos(A/2) = 2sin[(A+C)/2]cos[(A-C)/2]
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![已知△ABC和三内角A、B、C成等差数列,A为最小角,且√3cosA/2=sinA+sinC,求A、B、C的值.A,B,C成等差数列 ==> A+C=2B而 A+B+C =180°∴ A+B+C = 3B=180 ==> B=60°√3cos(A/2) = sinA+sinC==>√3cos(A/2) = 2sin[(A+C)/2]cos[(A-C)/2]](/uploads/image/z/8578257-33-7.jpg?t=%E5%B7%B2%E7%9F%A5%E2%96%B3ABC%E5%92%8C%E4%B8%89%E5%86%85%E8%A7%92A%E3%80%81B%E3%80%81C%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2CA%E4%B8%BA%E6%9C%80%E5%B0%8F%E8%A7%92%2C%E4%B8%94%E2%88%9A3cosA%2F2%3DsinA%2BsinC%2C%E6%B1%82A%E3%80%81B%E3%80%81C%E7%9A%84%E5%80%BC.A%2CB%2CC%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%3D%3D%3E+A%2BC%3D2B%E8%80%8C+A%2BB%2BC+%3D180%C2%B0%E2%88%B4+A%2BB%2BC+%3D+3B%3D180+%3D%3D%3E+B%3D60%C2%B0%E2%88%9A3cos%28A%2F2%29+%3D+sinA%2BsinC%3D%3D%3E%E2%88%9A3cos%28A%2F2%29+%3D+2sin%5B%28A%2BC%29%2F2%5Dcos%5B%28A-C%29%2F2%5D)
已知△ABC和三内角A、B、C成等差数列,A为最小角,且√3cosA/2=sinA+sinC,求A、B、C的值.A,B,C成等差数列 ==> A+C=2B而 A+B+C =180°∴ A+B+C = 3B=180 ==> B=60°√3cos(A/2) = sinA+sinC==>√3cos(A/2) = 2sin[(A+C)/2]cos[(A-C)/2]
已知△ABC和三内角A、B、C成等差数列,A为最小角,且√3cosA/2=sinA+sinC,求A、B、C的值.
A,B,C成等差数列 ==> A+C=2B
而 A+B+C =180°
∴ A+B+C = 3B=180 ==> B=60°
√3cos(A/2) = sinA+sinC
==>√3cos(A/2) = 2sin[(A+C)/2]cos[(A-C)/2]
只需解释下最后一步是怎么得来的就好了,
已知△ABC和三内角A、B、C成等差数列,A为最小角,且√3cosA/2=sinA+sinC,求A、B、C的值.A,B,C成等差数列 ==> A+C=2B而 A+B+C =180°∴ A+B+C = 3B=180 ==> B=60°√3cos(A/2) = sinA+sinC==>√3cos(A/2) = 2sin[(A+C)/2]cos[(A-C)/2]
和差化积公式
设公差为α
√3cos(A/2) = 2sin[(A+C)/2]cos[(A-C)/2]
√3cos(A/2) = 2sin60cos[(2α)/2]
√3cos(A/2) = √3cosα
所以A=2α
因为A+B+C=2α+60+60+α=180
解得α=20°
所以A=40°,B=60°,C=80°