A large share of such questions are addressed via known sum of series. With trigonometrics arguments like $\cos(\alpha n+\beta)$, one solution is to use Euler/De Moivre formulae:
$$\cos(\alpha n+\beta) = \frac{e^{j(\alpha n+\beta)}+e^{-j(\alpha n+\beta)}}{2}$$
and recognize, under the Fourier sum, two geometric series:
$$\sum_{n=0}^{N-1} \frac{e^{j(\alpha n+\beta)}+e^{-j(\alpha n+\beta)}}{2}e^{-j2\pi kn/N}$$
with terms like :
$$\sum_{n=0}^{N-1}\left( e^{j(\pm\alpha -j2\pi k/N)}\right)^n$$
whose sum is well-known. For a given $k$, you have:
$$
\begin{align}
\sum_{n=0}^{N-1} e^{\pm j(\alpha n+\beta)}e^{-j2\pi kn/N} &= & e^{\pm j\beta}\sum_{n=0}^{N-1} e^{\pm j\alpha n-j2\pi kn/N}\\
&= & e^{\pm j\beta}\sum_{n=0}^{N-1} e^{j(\pm \alpha -2\pi k/N)n}\\
&= & e^{\pm j\beta}\sum_{n=0}^{N-1} \left(e^{j(\pm \alpha -2\pi k/N)}\right)^n\\
\end{align}
$$
and then, noting
$$ \rho = e^{j(\pm \alpha -2\pi k/N)}$$
you can get:
$$\sum_{n=0}^{N-1}\rho^n =\begin{cases}N \textrm{ when } \rho=1, \\\frac{1-\rho^N}{1-\rho} \textrm{ when } \rho\neq 1.\end{cases} $$
