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My classmate asked a question during lecture about our discussion of bounded sequences, particularly the sequence $\sin(n)$. His question was, What is $\sup(\sin(n))$?

Secure Space
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Just to complete @AndreNicolas' argument, you want to prove the following : Let $a_n = n\text{mod}(2\pi)$, and fix $x_0 = \pi/2 \in [0,2\pi]$. For any $\delta > 0$, there exists $n\in \mathbb{N}$ such that $$ |a_n-x_0| < \delta $$ Proof : Choose $N \in \mathbb{N}$ such that $1/N < \delta/2\pi$, and consider the set $$ \{a_0, a_1, \ldots, a_N\} $$ Note that these points are distinct, because if $a_k = a_l$, then $(k-l) \equiv 0\text{mod}(2\pi)$, which would imply that $\pi$ is rational, which it isn't.

Now consider the intervals $$ \left[0,\frac{2\pi}{N}\right], \left[\frac{2\pi}{N}, \frac{4\pi}{N}\right], \ldots,\left[\frac{2\pi(N-1)}{N},2N\right] $$ There are $N$ such intervals and $(N+1)$ points. Hence, by the Pigeon-Hole principle, two such points must lie inside a single sub-interval. i. e., there exists $0\leq l < k \leq N$ such that $$ |a_k - a_l| = (k-l)\text{mod}(2\pi) < 2\pi/N < \delta $$ Let $\alpha = (k-l)\text{mod}(2\pi)$, and choose the smallest natural number $m$ such that $$ m\alpha > x_0 $$ (This is possible since $\alpha > 0$, and hence $m\alpha \to \infty$. Now since $m$ is minimal, $(m-1)\alpha < x_0$. Thus $$ a_{(m-1)(k-l)} < x_0 < a_{m(k-l)} $$

Hence, for $n=m(k-l)$, we have $$ |a_n-x_0| < a_{m(k-l)} - a_{(m-1)(k-l)} = \alpha < \delta $$

So to complete your argument, use the continuity of $\sin(x)$ at $x=\pi/2$ : For any $\epsilon > 0$, there exists $\delta > 0$ such that $$ |x-x_0| < \delta \Rightarrow |\sin(x) - 1| < \epsilon $$ For this delta, there exists $n\in \mathbb{N}$ such that $|a_n-x_0| < \delta$. Hence, $$ |\sin(n) -1| = |\sin(a_n) - 1| <\epsilon $$ Thus $\sup(\sin(n)) = 1$

Moreblue
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Prahlad Vaidyanathan
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The set of points of the form $\sin(n)$, where $n$ ranges over the integers, is bounded above by $1$. It turns out that $1$ is the supremum of the set.

One can show that the numbers of the form $\sin(n)$ are in fact dense in the interval $[-1,1]$. So for any number $x$ in that interval, there are integers $n$ such that $\sin(n)$ is arbitrarily close to $x$.

There is no integer $n$ such that $\sin(n)=1$. However, there is an integer $n$ such that $\sin(n)\gt 0.99$. There is also an integer $n$ such that $\sin(n)\gt 0.9999$. And so on.

André Nicolas
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  • can you provide a proof of that statement? – roger Sep 04 '13 at 17:12
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    The proof that $\sin n$ is never equal to $1$ takes a good deal of work, it is no easier I think than the result that $\pi$ is not rational. The proof that $\sin(n)$ is dense has been done several times on MSE, probably at least once by me. It uses the irrationality of $\pi$. To me it is easiest to use complex numbers, and consider $e^{in}$. By irrationality of $\pi$, we have if $m$ and $n$ are distinct positive integers, $e^{im}\ne e^{in}$. By Pigeonhole, for any $\epsilon\gt 0$, there are $m\ne n$ such that the argument of $e^{im}$ (Cont.) – André Nicolas Sep 04 '13 at 17:22
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    (Conr.) differs from the argument of $e^{in}$ by less than $\epsilon$. Then by taking integer powers of $e^{i(n-m)}$ we can get $\epsilon$-close to any point on the unit circle. – André Nicolas Sep 04 '13 at 17:24