In a Banach space $x_n\text{cos}(nt)+ y_n\text{sin}{nt}\rightarrow 0$ implies both sequences individually go to $0$. Let $X$ be a Banach space, and let $\{x_n\}$, $\{y_n\}$ be two sequences in $X$. Suppose $x_n\text{cos}(nt)+ y_n\text{sin}(nt)\rightarrow 0$ as $n \rightarrow \infty$ for all $t$ in a non degenerate interval. Show that $\{x_n\}$ and $\{y_n\}$ both go to $0$.
I'm trying to show that both the sequences go to $0$ along each subsequence or at least show that each subsequence has a further subsequence along which both go to $0$. To do this, I thought of somehow choosing the further subsequence such that for some $t$ in the given interval, $e^{int}$ is quite close to $1$ along the further subsequence which should give me information about  $\{x_n\}$. However, irrational rotations on the circle equidistribute which makes me clueless as to how to achieve this. 
 A: I think i found a solution using your strategy (i.e. choosing a subsequence such that $e^{int}$ is close to $1$). However, we choose a subsequence such that $e^{int}$ is close to an arbitrary point $\alpha$ on the unit circle.
Let $M\subset\mathbb{N}$ be the index set of an arbitrary subsequence (that is, with $\sup M=\infty$).
We denote the interval containing the variables $t$ by $(l_0,u_0)$.
For a large $n_1\in M$ we can find $s_1\in (l_0,u_0)$ such that $e^{in_1s_1}=\alpha$.
Due to continuity,
we can select $l_1,u_1$ around $s_1$ such that $e^{in_1t}$ is close to $\alpha$ for all $t\in (l_1,u_1)$.
To be specific, we require
$$
|l_1-u_1|<\frac12 \qquad\text{and}\qquad
|e^{in_1t}-\alpha|<\frac12 \quad\forall t\in(l_1,u_1).
$$
We repeat this process to recursively define our subsequence:
Assume that $n_{k-1}, l_{k-1},u_{k-1}$ are already constructed.
For a large $n_k\in M, n_k>n_{k-1}$ we can find $s_k\in (l_{k-1},u_{k-1})$ such that $e^{in_ks_k}=\alpha$.
Due to continuity,
we can select $l_k,u_k$ around $s_k$ such that
$$
|l_k-u_k|<2^{-k} \qquad\text{and}\qquad
|e^{in_kt}-\alpha|<2^{-k} \quad\forall t\in(l_k,u_k).
$$
Now that we have the sequence constructed, it can be seen that $l_k$ and $u_k$ converge to the same point,
which we call $r$.
Since $l_k<r<u_k$ for all $k$, it follows that $e^{in_kr}\to \alpha$ as $k\to\infty$.
To summarize, we have shown, that for each subsequence (here denoted by the index set $M$) and each $\alpha$ on the unit circle, we can find $r$ such that $e^{in_kr}\to \alpha$.
In order to apply this, suppose that $M_1\subset\mathbb{N}$ is given such that $\|x_n\|\geq \varepsilon$ for all $n\in M_1$.
Using the above with $\alpha=1$, it can be seen that $\|y_{n_k}\|\|x_{n_k}\|^{-1} \to\infty$ for a suitable subsequence.
If we denote the indices of this new subsequence by $M_2\subset M_1$,
we can apply the same result using $\alpha=i$.
This yields a subsequence such that $\|x_{n_k}\|\|y_{n_k}\|^{-1}\to\infty$.
Because this subsequence is a subsequence of the former subsequence, this is a contradiction to $\|y_{n_k}\|\|x_{n_k}\|^{-1} \to\infty$.
Therefore $x_n$ converges to zero. Another application of the auxilliary result with $\alpha=i$ yields the same for $y_n$.
