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(a) We want to find odd multiples of $\pi$ that are close to integers.

• Solution 1: One way to do that is to remember that $\pi$ is somewhat close to $\dfrac{22}{7}$. Then when we multiply $\pi$ by a multiple of 7, we should get something close to an integer. In particular, $7\pi$, $21\pi$, and $35\pi$ should be reasonably close to $7\left(\dfrac{22}{7}\right)=22$, $21\left(\dfrac{22}{7}\right)=66$, and $35\left(\dfrac{22}{7}\right)=110$, respectively. We check whether they are close enough:

$$\begin{aligned} 7\pi&\approx 21.99 &\qquad 21\pi &\approx 65.97 & \qquad 35\pi &\approx 109.96 \end{aligned}$$

So indeed, $22$, $66$, and $110$ are all within 0.1 of some odd multiple of $\pi$.

Since the cosine of an odd multiple of $\pi$ is $-1$, we expect all of the sequence values to be close to $-1$. Using a calculator:

$$\begin{aligned}a_{22} &= \cos(22) \approx -0.99996,\\ a_{66} &= \cos(66) \approx -0.99965, \\ a_{110} &= \cos(110) \approx -0.99902 \end{aligned}$$

• Solution 2: Alternately, we could have just listed odd multiple of $\pi$ until we found three that are close to integers. $$\begin{array}{c|c} \hline \mathbf{2k+1}&\mathbf{(2k+1)}\boldsymbol{\pi}\\ \hline 1&3.14\\ 3&9.42\\ 5&15.71\\ 7&21.99\\ 9&28.27\\ 11&34.56\\ 13&40.84\\ 15&47.12\\ 17&53.41\\ 19&59.69\\ 21&65.97\\ 23&72.26\\ 25&78.54\\ 27&84.82\\ 29&91.11\\ 31&97.39\\ 33&103.67\\ 35&109.96 \end{array}$$ Some earlier odd multiples of $\pi$ (like $15\pi$ and $29\pi$) get fairly close to integers, but not within 0.1.

(b) If $x = \dfrac{2k+1}{\vphantom{_1}2}\pi$ for some integer $k$ (that is, $x$ is an odd multiple of $\pi/2$), then $\cos x =0$. So, we can either list out the first few terms of $a_n$ until we find three that are very close to $0$, or we can use our approximation $\pi\approx \dfrac{22\vphantom{^1}}{7}$ to choose values of $n$ that are close to $\dfrac{2k+1}{2}\pi$.

• Solution 1: $$\begin{aligned} \dfrac{2k+1}{2}\pi&\approx \frac{(2k+1)\times 22}{2\times 7}=11\times\frac{2k+1}{7} \end{aligned}$$

So, we expect our values to be close to integers when $2k+1$ is a multiple of 7. For example, $2k+1=7$, $2k+1=21$, and $2k+1=35$.

We check: $$\begin{array}{l| l| l} \qquad\mathbf{x}&\mathbf{n} & \mathbf{a_n}\\ \hline ~7\times \dfrac{\pi\vphantom{^A}}{2}\approx 10.99557 & 11&a_{11}\approx0.0044\\[10pt] 21\times \dfrac{\pi}{2} \approx 32.98672&33&a_{33}\approx-0.0133\\[10pt] 35\times \dfrac{\pi}{2} \approx 54.97787&55& a_{55}\approx 0.0221\\ \end{array}$$ These seem like values of $a_n$ that are all pretty close to 0.

• Solution 2: We could have listed the first several values of $a_n$, and looked for some that are close to 0.

$$\begin{array}{c|c} \mathbf{n}&\mathbf{a_n}\\ \hline 1&0.54\\ 2&-0.42\\ 3&-0.99\\ 4&-0.65\\ 5&0.28\\ 6&0.96\\ 7&0.75\\ 8&-0.15\\ 9&-0.91\\ 10&-0.84 \end{array}$$ Oof. Nothing very close yet. Maybe a better way is to list values of $\frac{2k+1}{2}\pi$, and see which ones are close to integers.

$$\begin{array}{c|c} \mathbf{2k+1}&\mathbf{\frac{2k+1}{2}}\boldsymbol{\pi}\\ \hline 1&1.57\\ 3&4.71\\ 5&7.85\\ 7&10.996\\ 9&14.14\\ 11&17.28\\ 13&20.42\\ 15&23.56\\ 17&26.70\\ 19&29.85\\ 21&32.99\\ 23&36.13\\ 25&39.27\\ 27&42.41\\ 29&45.55\\ 31&48.69\\ 33&51.84\\ 35&54.98 \end{array}$$ We find roughly the same candidates we did in Solution 1, depending on what we're ready to accept as “close".

(c) One can use the same strategies as we did for parts (a) and (b). But since we already know some $n$'s with $\cos(n)$ close to $-1$, it's easier to use the trig identity

$$\begin{aligned} \cos(2m) = 2\cos^2(m) -1 \end{aligned}$$

This identity shows that if $\cos(m)$ is close to $-1$, then $\cos(2m)$ is close to $+1$. So let's try $n=2m = 2\times 22=44$, $2\times 66=132$ and $2\times 110=220$. $$\begin{array}{c|c} \mathbf{n}&\mathbf{a_n}\\ \hline 44&0.9998\\ 132&0.9986\\ 220&0.9961 \end{array}$$ They do the trick.

Remark: it is possible to turn the ideas of this question into a rigorous proof that $\lim\limits_{n \to \infty}\cos n$ is undefined.

• Let, for each integer $k\ge 1$, $n_k$ be the integer that is closest to $2k\pi$. Then $2k\pi-\frac{1}{2}\le n_k \le 2k\pi+\frac{1}{2}$ so that $\cos(n_k)\ge\cos\frac{1}{2}\ge 0.8$. Consequently, if $\lim\limits_{n \to \infty}\cos n=c$ exists, we must have $c\ge 0.8$.
• Let, for each integer $k\ge 1$, $n'_k$ be the integer that is closest to $(2k+1)\pi$. Then $(2k+1)\pi-\frac{1}{2}\le n'_k \le (2k+1)\pi+\frac{1}{2}$ so that $\cos(n'_k)\le-\cos\frac{1}{2}\le -0.8$. Consequently, if $\lim\limits_{n \to \infty}\cos n=c$ exists, we must have $c\le -0.8$.
• It is impossible to have both $c\ge 0.8$ and $c\le -0.8$, so $\lim\limits_{n \to \infty}\cos n$ does not exist.

Remark: This question also hints at a property of the set of all numbers $\cos(n)$, with $n$ running over the integers. Mathematicians say that this set is “dense in the interval $[-1,1]$”. This means that if you pick any number $-1\le r\le 1$, there is an integer $n$ such that $\cos(n)$, while not necessarily being exactly $r$, is as close to $r$ as you like. If you feed “$\cos(n)$ dense” into your favourite search engine, you can find out more.