Talk:Cosine of Integer Multiple of Argument/Formulation 6

Is there a particular reason why this result (and the presentation of its proof) was expressed with the negative terms first?

I appreciate that it's not incorrect as such, just that it seems a bit clumsy. It's easier to grasp expressed as:

$\cos n \theta = \map \cos {\paren {n - 2} \theta} - 2 \sin \theta \map \sin {\paren {n - 1} \theta}$

Any reason for doing it the way round you did it?

--prime mover (talk) 07:07, 25 February 2021 (UTC)

Consistency. My intention was to consistently express all of the recursive sequences in a similar fashion:

$\cos n \theta = $ Something x prior term + prior to the prior term

That $-2 \sin \theta$ term is the star of the show in the continued fraction expansion

--Robkahn131 (talk) 04:20, 26 February 2021 (UTC)