N-rowed determinant which corresponds to cosine multiple angle formula

[Seydlitz]

Homework Statement

I need to prove or show that this n-rowed determinant which corresponds to cosine multiple angle formula is in fact true using induction.

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

 

[PeroK]

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

 

[Seydlitz]

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

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