Ross' handwaving can be justified.
We use Weyl's theorem on the equidistribution of polynomials (mod 1) with at least one irrational coefficient (see here: How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1?)
Choose $\displaystyle p(x) = 2\pi x + \pi$.
Let $\displaystyle k \in (0,2)$. Let $\displaystyle \delta = k$ if $\displaystyle k \in (0,1)$ and $\displaystyle \delta = k-1$ otherwise.
By Weyl's theorem, density of $n$ such that the fractional part of $\displaystyle p(n) < 1- \delta$ is non zero.
Thus the density (among the set of natural numbers) of the odd numbers $\displaystyle 2n+1$ for which the fractional part of $\displaystyle (2n+1)\pi < 1 - \delta$ is non-zero. Thus the density of $\displaystyle [(2n+1)\pi]$ is non-zero too.
For such an $\displaystyle n$,
Let $\displaystyle N = [(2n+1)\pi]$.
We have that $\displaystyle (2n+1)\pi - N < 1 - \delta$ and so $\displaystyle \cos(N) - \cos((2n+1)\pi) < 1 - \delta$
And thus $\displaystyle \cos(N) < -\delta$ and so $\displaystyle \cos(N) + k < k - \delta \leq 1$
Thus for we have a positive density subset $\displaystyle S$ of $\displaystyle \mathbb{N}$ for which $\displaystyle \cos(N) + k < 1$.
We have that
$$ \sum_{N \in S} \frac{1}{N^{k+\cos(N)}} > \sum_{N \in S} \frac{1}{N}$$
Since $\displaystyle S$ is of positive density, the sum of reciprocals is divergent, for a proof see here: Theorem on natural density
I will leave my earlier attempt here.
For $k \in (0,1)$ consider the below.
Lemma:
Given any $\displaystyle k \in (0,1)$, we can find an infinite number of odd numbers $\displaystyle M$ such that $\displaystyle M\pi - [M\pi] < 1 - k$.
Proof:
This easily follows from Vinogradov's theorem that $\displaystyle \\{p_{n}\alpha\\}$ is equidistributed for irrational $\displaystyle \alpha$ where $\displaystyle p_{n}$ is the $\displaystyle n^{th}$ prime. (See here: http://en.wikipedia.org/wiki/Equidistributed_sequence)
$\displaystyle \circ$
For $\displaystyle n = [M\pi]$ we have
$\displaystyle \cos(n) - \cos(M\pi) = \cos(n) + 1 < 1- k$
(using $\displaystyle |\cos x - \cos y| < |x-y|$).
And so $\displaystyle \cos(n) + k < 0$ for an infinite number of $\displaystyle n$ and so the series $\displaystyle \sum \frac{1}{n^{k+\cos n}}$ must be divergent.
For the $\displaystyle \sin n$ case, we need the sequence $\displaystyle \\{(4k+3)\pi/2\\}$ which is an infinite subsequence of $\displaystyle \\{p_{n}\pi/2\\}$ and so the series $\displaystyle \sum \frac{1}{n^{k+\sin n}}$ is divergent.