Problems about $\sin(n)$ where $n$ is an integer $\in (0,1000]$

Question

Let $n$ be an integer such that $0<n \le 1000$. Find $n$ so that:

• $\sin(n)$ is minimized.
• $\sin(n)$ is maximized.
• $|\frac{\sqrt{2}}{2}-\sin(n)|$ is minimized.

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Which is larger; $\sin(710)$ or $-\sin(355)$?

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Using MicrosoftExcel, I found that:

• $\sin(n)$ is minimized when $n=344$.

• $\sin(n)$ is maximized when $n=699$.

• $|\frac{\sqrt{2}}{2}-\sin(n)|$ is minimized when $n=882$.

• $\sin(710)>-\sin(355)$

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Can we answer those four questions without calculators/Excel/...?

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Any help would be appreciated. THANKS!

Answer 1

There's no clever way to do this by hand. It's probably intended for you to use a program.

The point of the exercise is to illustrate how close you can get to particular values using integer inputs. In fact you can get arbitrarily close (without being equal) to any value between $-1$ and $1$ using integer inputs.

Answer 2

It's equivalent to: $$\min_{k} |(4k+r)\pi-n|$$ For first one, $r=-\frac{1}{2}$. And $r=0,2$ if you mean its absolute value.

For second one, $r=\frac{1}{2}$. And $r=\pm\frac{1}{2}$ if you mean its absolute value.

For Third one, $r=\frac{1}{4},\frac{3}{4}$. And $r=\pm\frac{1}{4},\pm\frac{3}{4}$ if summation also considered.

For Forth one, by radian degree unit assumption: $$-\sin(355)=-\sin(113\pi+\frac{\epsilon}{2})=\sin(\frac{\epsilon}{2}) < \sin(\epsilon)=\sin(226\pi+\epsilon)=\sin(710) $$ That it follow from $0 < \epsilon < \frac{\pi}{2}$ and increasing of $\sin$ on [$-\frac{\pi}{2},+\frac{\pi}{2}]$. For other degree unit the same method should be applied. By similar argument you can see, in fact, if you consider the absolute value, so $n=344$ become your $\max$.

Answer 3

We will use identity $$\sin(n)=\sin\left(n-2\pi k\right),$$ where $k\in\mathbb{Z}\;$ so that $$ n-2\pi k \in (-\pi; \pi]. $$

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The case "min": we focus on expressions $$ n-2\pi k = -\dfrac{\pi}{2}+\varepsilon, $$ $$ 2n-(4k-1)\pi = 2\varepsilon, $$ $$ \dfrac{2n}{4k-1} - \pi = \delta,\tag{1} $$ where $\delta = \dfrac{2\varepsilon}{4k-1}$, and $\varepsilon$ is as small as possible (by absolute value). Other words, $$ \dfrac{2n}{4k-1}\tag{2} $$ has to be better approximation of $\large{\pi}$ among fractions of the form $(2)$.

It's known that better (at all) approximations are convergents of appropriate continued fraction (the larger numerator $-$ the better approximation).

Widely known that $\pi$ continued fraction has these convergents:
$$\dfrac{3}{1}, \; \dfrac{22}{7}, \; \dfrac{333}{106}, \; \dfrac{355}{113}, \; \dfrac{103993}{33102}, ... $$

We will construct the answer from them. When consider fractions with numerators between $355$ and $103993$, then those ones are (maybe some rigorous proof is needed here) fractions of the form: $$ \dfrac{333+355s}{106+113s},\tag{3} $$ while $s=1,2,...,292$.

We consider just $s=1,2,3,4$ (remember on limitation for $n$). Only $s=1$ matches the form $(2)$. Easy to see that $n=344$ in this case.

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The case "max": same way, $$ n-2\pi k = \dfrac{\pi}{2}+\varepsilon, $$ $$ \dfrac{2n}{4k+1} - \pi = \delta,\tag{4} $$ where $\delta = \dfrac{2\varepsilon}{4k+1}$. We focus on $\pi$-approximations of the form $$ \dfrac{2n}{4k+1}.\tag{5} $$

We will focus on $s=1,2,3,4$ (remember on limitation for $n$). Only $s=3$ (see $(3)$) matches the form $(5)$. Easy to see that $n=699$ in this case.

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The case $\left|\dfrac{\sqrt{2}}{2} - \sin(n)\right|$: similar way: $$ n-2\pi k = \dfrac{\pi}{4}+\varepsilon, $$ $$ 4n-(8k+1)\pi = 4\varepsilon, $$ $$ \dfrac{4n}{8k+1} - \pi = \delta,\tag{6} $$ where $\delta = \dfrac{4\varepsilon}{8k+1}$. We focus on $\pi$-approximations of the form $$ \dfrac{4n}{8k+1}.\tag{7} $$ Candidates for $s$ are now: $s=1,2,3,4,5,6,7,8,9,10$. Only $s=1,9$ (see $(3)$) match the form $(7)$. But $s=9$ provides larger numerator (better approximation). Easy to see that $n=882$ in this case.