在三角ABC中,a,b,c分别是角A,角B,角C的对边.求证:acos^2 B/2+bcos^2 A/2=1/2(a+b+c)RT
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![在三角ABC中,a,b,c分别是角A,角B,角C的对边.求证:acos^2 B/2+bcos^2 A/2=1/2(a+b+c)RT](/uploads/image/z/12079398-30-8.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92ABC%E4%B8%AD%2Ca%2Cb%2Cc%E5%88%86%E5%88%AB%E6%98%AF%E8%A7%92A%2C%E8%A7%92B%2C%E8%A7%92C%E7%9A%84%E5%AF%B9%E8%BE%B9.%E6%B1%82%E8%AF%81%EF%BC%9Aacos%5E2+B%2F2%2Bbcos%5E2+A%2F2%3D1%2F2%28a%2Bb%2Bc%29RT)
在三角ABC中,a,b,c分别是角A,角B,角C的对边.求证:acos^2 B/2+bcos^2 A/2=1/2(a+b+c)RT
在三角ABC中,a,b,c分别是角A,角B,角C的对边.求证:acos^2 B/2+bcos^2 A/2=1/2(a+b+c)
RT
在三角ABC中,a,b,c分别是角A,角B,角C的对边.求证:acos^2 B/2+bcos^2 A/2=1/2(a+b+c)RT
2[acos^2 B/2+bcos^2 A/2]
=2[a(cosB+1)/2+b(cosA+1)/2]
=acosB+bcosA+a+b
=a*(a^2+c^2-b^2)/(2ac)+b(b^2+c^2-a^2)/(2bc)+(a+b)
=(a^2+c^2-b^2+b^2+c^2-a^2)/(2c)+(a+b)
=2c^2/(2c)+(a+b)
=a+b+c
即:
2[acos^2 B/2+bcos^2 A/2]=a+b+c
所以:
acos^2 B/2+bcos^2 A/2=1/2(a+b+c)