# N-rowed determinant which corresponds to cosine multiple angle formula

1. Nov 27, 2013

### Seydlitz

1. The problem statement, all variables and given/known data
I need to prove or show that this n-rowed determinant which corresponds to cosine multiple angle formula is in fact true using induction.

3. The attempt at a solution
First let $a = \cos \theta$ and suppose I have this n by n determinant.

$$\begin{vmatrix} &a &1 &0& \\ &1 &2a &1& \\ &0 &1 &2a& \\ \end{vmatrix} = \cos n\theta$$

For $n=1$ it's easy to see that the determinant does in fact correspond to $\cos n\theta$.

I assume it's true for $n$, and I want to show that it's the case for $n+1$ to complete the induction. So this is the new n+1 by n+1 determinant.

$$\begin{vmatrix} &a &1 &0 &0&\\ &1 &2a &1 &0&\\ &0 &1 &2a &1&\\ &0 &0 &1 &2a&\\ \end{vmatrix}= \cos (n+1)\theta$$

Evaluating the determinant using the last column I can have.
$$2 \cos \theta \begin{vmatrix} &a &1 &0& \\ &1 &2a &1& \\ &0 &1 &2a& \\ \end{vmatrix}- \begin{vmatrix} &a &1 &0& \\ &1 &2a &1& \\ &0 &0 &1& \\ \end{vmatrix}$$

The first term is equal to $2 \cos \theta \cos n\theta$.

My problem is to relate the second term to the sine term in this angle identity.

$\cos (n+1)\theta = \cos \theta \cos n\theta - \sin \theta \sin n\theta$.

If that's done then the prove should be done too.

Thanks

2. Nov 27, 2013

### PeroK

Hint: that second term looks like a previous term in the sequence!

3. Nov 27, 2013

### Seydlitz

Got it thanks! What a powerful hint you have there. :)