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N-rowed determinant which corresponds to cosine multiple angle formula

  1. Nov 27, 2013 #1
    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. jcsd
  3. Nov 27, 2013 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: that second term looks like a previous term in the sequence!
     
  4. Nov 27, 2013 #3
    Got it thanks! What a powerful hint you have there. :)
     
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