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
證明:行列式記為Dn.
按第1列展開得:Dn=2cosθD(n-1) - D(n-2).
下用歸納法證明
當n=1時,D1=2cosθ
sin(n+1)θ/sinθ=sin2θ/sinθ=2cosθ.
所以n=1時結論成立,即D1=sin(1+1)θ/sinθ.
假設k
高數(shù)求證行列式
高數(shù)求證行列式
2cosa1 0 0 0``````00
1 2cosa1 0 0 `````` 00
0 12cosa 1 0 ```````00
0 0 12cosa 1````` 00
```````````````````
0 0 0 00 ``````2cosa 1
0 0 0 00 `````` 1 2cosa
=sin(n+1)a / sina
2cosa1 0 0 0``````00
1 2cosa1 0 0 `````` 00
0 12cosa 1 0 ```````00
0 0 12cosa 1````` 00
```````````````````
0 0 0 00 ``````2cosa 1
0 0 0 00 `````` 1 2cosa
=sin(n+1)a / sina
數(shù)學人氣：300 ℃時間：2020-06-07 10:38:12
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