设ABC是三角形ABC的三个内角,若向量m=(1-cos(A+B)),n=(5/8,cos(A+B)/2)且m.n=9/8求证:tanA*tanB=1/9求absinC/(a²+b²-c²)的最大值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 05:16:58
![设ABC是三角形ABC的三个内角,若向量m=(1-cos(A+B)),n=(5/8,cos(A+B)/2)且m.n=9/8求证:tanA*tanB=1/9求absinC/(a²+b²-c²)的最大值](/uploads/image/z/5202101-29-1.jpg?t=%E8%AE%BEABC%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E8%8B%A5%E5%90%91%E9%87%8Fm%3D%EF%BC%881-cos%EF%BC%88A%2BB%EF%BC%89%EF%BC%89%2Cn%3D%EF%BC%885%2F8%2Ccos%28A%2BB%29%2F2%29%E4%B8%94m.n%3D9%2F8%E6%B1%82%E8%AF%81%EF%BC%9AtanA%2AtanB%3D1%2F9%E6%B1%82absinC%2F%EF%BC%88a%26%23178%3B%2Bb%26%23178%3B-c%26%23178%3B%EF%BC%89%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
设ABC是三角形ABC的三个内角,若向量m=(1-cos(A+B)),n=(5/8,cos(A+B)/2)且m.n=9/8求证:tanA*tanB=1/9求absinC/(a²+b²-c²)的最大值
设ABC是三角形ABC的三个内角,若向量m=(1-cos(A+B)),n=(5/8,cos(A+B)/2)且m.n=9/8
求证:tanA*tanB=1/9
求absinC/(a²+b²-c²)的最大值
设ABC是三角形ABC的三个内角,若向量m=(1-cos(A+B)),n=(5/8,cos(A+B)/2)且m.n=9/8求证:tanA*tanB=1/9求absinC/(a²+b²-c²)的最大值
m=(1-cos(A+B),cos (A-B)/2),n=(5/8,cos(A-B/2),mn=9/8
求证:tanA*tanB=1/9
求absinC/(a²+b²-c²)的最大值
【解】
mn=[1-cos(A+B)]*[5/8]+ cos ²[(A-B)/2]
=5/8-5/8* cos(A+B)+1/2*(1+cos(A-B))
=9/8-5/8* cos(A+B) +1/2* cos(A-B)
=9/8-5/8*(cosAcosB-sinAsinB) +1/2*(cosAcosB+sinAsinB)
=9/8-1/8* cosAcosB+9/8* sinAsinB,
因为mn=9/8
所以-1/8* cosAcosB+9/8* sinAsinB=0,
1/8* cosAcosB=9/8* sinAsinB
tanA*tanB=1/9.
tan(A+B)=( tanA+tanB)/(1- tanA*tanB)
=( tanA+tanB)/(1- 1/9)
=9/8*( tanA+tanB)……利用基本不等式
≥9/8*2√(tanA*tanB)
=(9/8)*2*(1/3)=3/4.
即-tanC≥3/4.tanC≤-3/4.
根据余弦定理知:cosC=(a²+b²-c²)/(2ab),
则a²+b²-c²=(2ab) *cosC
所以absinC/(a²+b²-c²)
=absinC/[(2ab) *cosC]
=1/2* tanC≤-3/8.