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

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 $$z$$.

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 $$\{\cos j\}$$ $$\blacksquare$$