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Prove that

  1. The sequence $y_n=\tan(n)$ has a convergent subsequence.

  2. $x_n=2\sin^3(n)+6\cos^5(2n)$ has a convergent subsequence.

For the second one, I need help showing it is bounded so that I can apply Bolzano-Weierstrass Theorem that every bounded sequence has a convergent subsequence.

Then for the first one, I can't apply this theorem since tangent is unbounded so I'm not sure how to approach this.

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For the first problem: tangent may be unbounded, but tangent restricted to angles between $-\pi/4$ and $\pi/4$ is bounded between $-1$ and $1$. So can you find a subsequence of $\langle 0,1,2,\dots\rangle$ such that the radian measure of those integers is always equivalent to angles between $-\pi/4$ and $\pi/4$?

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Hint for 1: First create a bounded subsequence. Pick an upper bound (say $10$ for example) and only choose $n_k \in \Bbb{N}$ such that $\left|\tan(n_k)\right| \leq 10$. Now $\left\{\tan(n_k) \right\}_{k=1}^\infty$ is a bounded sequence.

Hint for 2: $$\left|x_n \right|=\left|2\sin^3(n)+6\sin^5(n)\right| \\ \leq \left|2\sin^3(n)\right|+\left|6\sin^5(n)\right|$$ and $\left|\sin^m(n)\right| \leq 1$ for all $n,m \in \Bbb{N}$.

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Assuming that $\frac{p}{q}$ is a convergent of the continued fraction of $2\pi$, then $\sin(p),\tan(p)$ are close to zero and $\cos(2p)$ is close to one. Then you can apply Bolzano-Weierstrass theorem.

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