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Let $\theta_{kl}$ be an angle such that $\cos\theta_{kl}=\frac{1}{2}(\cos(\frac{2\pi k}{n})+\cos(\frac{2\pi l}{n}))$.

Given that definition, if I introduce a new variable $t$ is the following a correct?

$\cos(t\theta_{kl})\approx\frac{1}{2}(\cos(\frac{2\pi kt}{n})+\cos(\frac{2\pi lt}{n}))$

Update: I'm actually interested in the asymptotics of $\theta_{kl}$. By a second order approximation $\theta_{kl}^2=O(\frac{k^2+l^2}{n^2})$. Is it correct? If it is, then the above holds, right? But $\theta_{kl}$ needs to be small.

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There isn't any simple approximation of the sort you seek. Simply defining an angle $\theta_{k\ell}$ such that $\cos(\theta_{k\ell}) = \tfrac12 \bigl( \cos(\tfrac{2\pi k}n) + \cos(\tfrac{2\pi \ell}n) \bigr)$ is not sufficient. There are infinitely many different angles $\theta_{k\ell}$ which satisfy this equation, after all, which are very different from one another.

So what you want is to obtain some single value $\theta_{k\ell}$ which you can approximate well using rational multiples of a full rotation, as you describe. This will require mor than just a definition in terms of a periodic function such as cosine.

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My reasoning is the following:

By definition $\cos\theta_{kl}=\frac{1}{2}\cos\frac{2\pi k}{n}+\frac{1}{2}\cos\frac{2\pi l}{n}$.

Given that $\theta_{kl}$ is small, we take the Taylor expansion on both sides up to the second term and obtain

$1-\theta_{kl}^2\approx\frac{1}{2}(1-\frac{4\pi^2k^2}{n^2}+1-\frac{4\pi^2l}{n^2})=1-\frac{2\pi^2k^2+2\pi^2l^2}{n^2}$,

which implies

$\theta_{kl}=O(\frac{k+l}{n})$.