Proposition: The sequence $\{\cos j\}_{j=1}^\infty$ is dense in the interval $[-1,1]$.

Similar question, but uses more advance concepts:

What I have tried:

Idea 1: Pick an $\alpha \in [-1,1]$ and let $\cos^{-1}(\alpha)=j \in \Bbb{R}$. Consider the interval $(j-\delta,j+\delta)\subseteq \cos^{-1}([-1,1])$. By the density property of $\Bbb{Q}$, there exists $q \in (j-\delta, j+\delta)$; let $q=\frac{m}{n}$. Using the fact that $\operatorname{cosine}$ has a period of $2\pi$, we want to know what $k \in \Bbb{N}$ satisfies the inequality : $(j-\delta)+2\pi\cdot nk \lt n k\cdot \frac{m}{n} \lt (j+\delta)+2\pi\cdot nk$. Doing some manipulations, we get $\frac{j-\delta}{m-2\pi n} \lt k \lt \frac{j+\delta}{m-2\pi n}$. But, is there anyway that $k$ is a natural number?

Idea 2: Use the fact that $\operatorname{cosine}$ is continuous (But I don't think I can use it since I haven't reached continuity in the text yet.).Using it anyway, since $\operatorname{cosine}$ is continuous at $j$, we want to find $k,z\in\Bbb{N}$ so that $\vert j+2\pi k - z\vert \lt \delta$ which would imply $\vert \cos j -\cos z\vert \lt \epsilon$. What do you people think of this approach? Also, I'm stuck on how to find $k$ and $z$.


I elaborate the idea 2:

Let $t \in [-1,1]$. Since cosine function is continuous, and by Intermediate value theorem, there exist $x$ such that $$\cos x =t$$

I assume you know $\Bbb{Z}+2\pi\Bbb{Z}$ is dense in $\Bbb{R}$ [If you don't know see this post ]

Thus, there exist $x_n$ and $y_n$ such that $$x=\lim_{n \rightarrow \infty}(x_n+2\pi y_n)$$

Now, $$t=\cos x=\cos \Big(\lim_{n \rightarrow \infty}(x_n+2\pi y_n)\Big)=\lim_{n \rightarrow \infty}(\cos x_n)$$ where the last equality follows from the continuity of the cosine function

Hence each member of $[-1,1]$ is a limit point of $\{\cos j\}$ $\blacksquare$


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