- #1
Seydlitz
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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