![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
这道题目怎么做呀? 10
1个回答
展开全部
2.(1)由余弦定理
cosB=(a²+c²-b²)/2ac≥(2ac-b²)/2ac=1/2
又B∈(0,π) 所以B(0,π/3)
(2)y=(sin²B+cos²B+2sinBcosB)/(sinB+cosB)
=(sinB+cosB)²/(sinB+cosB)=sinB+cosB=√2sin(B+π/4)
0<B<=60
45<B+45<=105
所以sin(B+π/4)∈(√2/2,1]
所以y∈(1,√2]
cosB=(a²+c²-b²)/2ac≥(2ac-b²)/2ac=1/2
又B∈(0,π) 所以B(0,π/3)
(2)y=(sin²B+cos²B+2sinBcosB)/(sinB+cosB)
=(sinB+cosB)²/(sinB+cosB)=sinB+cosB=√2sin(B+π/4)
0<B<=60
45<B+45<=105
所以sin(B+π/4)∈(√2/2,1]
所以y∈(1,√2]
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询