高数求证行列式2cosa 1 0 0 0`````` 0 0 1 2cosa 1 0 0 `````` 0 00 1 2cosa 1 0 ``````` 0 00 0 1 2cosa 1 ````` 0 0```````````````````0 0 0 0 0 ``````2cosa 10 0 0 0 0 `````` 1 2cosa=sin(n+1)a / sina
问题描述：
高数求证行列式
2cosa 1 0 0 0`````` 0 0
1 2cosa 1 0 0 `````` 0 0
0 1 2cosa 1 0 ``````` 0 0
0 0 1 2cosa 1 ````` 0 0
```````````````````
0 0 0 0 0 ``````2cosa 1
0 0 0 0 0 `````` 1 2cosa
=sin(n+1)a / sina
答
记上述行列式为D(n)
按第一列展开
D(n)=2cosaD(n-1)+D(n-2)
D(1)=2COSa
然后可以用数学归纳法证明了。
答
按第一行展开
答
2cosθ 1 0 ...0 0
1 2cosθ 1 ...0 0
...
0 0 0 ...2cosθ 1
0 0 0 ...1 2cosθ
=sin(n+1)θ/sinθ
证明:行列式记为Dn.
按第1列展开得:Dn=2cosθD(n-1) - D(n-2).
下用归纳法证明
当n=1时,D1=2cosθ
sin(n+1)θ/sinθ=sin2θ/sinθ=2cosθ.
所以n=1时结论成立,即D1=sin(1+1)θ/sinθ.
假设k