solving a linear recurrence relationship involving trig

Question

can anyone gave me some idea of how to solve $a_{n+2}-2\cos(\theta)a_{n+1}+a_n=0$ ? I am not sure how to perform a calculation to this three term recurrence-- I know I can solve a two term recurrence but a characteristic polynomial but I really have no idea how to do this-- some help will be appreciated. Thanks!

Answer 1

As mentioned by GEdgar, the characteristic polynomial is $$ z^2 - 2 \cos(\theta)z + 1 \text{.} $$ It's worth pointing out that "$\cos \theta$" is just some constant, so you could replace it with some parameter and ignore its trigonometry-ness. The characteristic polynomial factors as $$ (z - (\cos \theta + \mathrm{i} \sin \theta))(z - (\cos \theta - \mathrm{i} \sin \theta)) \text{,} $$ so the general solution to the recurrence is $$ C_1(\cos \theta + \mathrm{i} \sin \theta)^n + C_2(\cos \theta - \mathrm{i} \sin \theta)^n \text{.} $$

We use the fact(s): $$ 1 = \cos^2(\theta) + \sin^2(\theta) = (\cos \theta + \mathrm{i} \sin \theta)(\cos \theta - \mathrm{i} \sin \theta) \text{.} $$