What is the least positive integer $n$ for which $|\sin(n)-\sin(n^{\circ})|<0.005$ What is the least positive integer $n$ for which $|\sin(n)-\sin(n^{\circ})|<0.005$?
(That is the difference between sine of $n$ radians and sine of $n$ degrees is less than $0.005$)
My attempt:


*

*$\sin(n^{\circ})=\sin(\frac{n\pi}{180})$

*$\sin(A)-\sin(B)=2\sin(\frac{A-B}{2})\cos(\frac{A+B}{2})$
Putting $A=n$, and $B=\frac{n\pi}{180}$, the given inequality becomes;
$$|2\sin(\frac{n-\frac{n\pi}{180}}{2})\cos(\frac{n+\frac{n\pi}{180}}{2})|<0.005$$
Simplifying, we get;
$$|\sin(\frac{n(180-\pi)}{360})\cos(\frac{n(180-\pi)}{360})|<0.0025$$
Substituting $u=\frac{n(180-\pi)}{360}$, we get;
$\sin(u)\cos(u)<0.0025$
Expanding both sine and cosine, we get:
$(u-\frac{u^3}{6}+\dots)(1-\frac{u^2}{2}+\dots)<0.0025$
or
$u-\frac{u^3}{2}-\frac{u^3}{6}+\frac{u^5}{12}-\dots<0.0025$
I stuck here to find $u$. Otherwise, by finding $u$, I can solve for $n$ since $n=\frac{360u}{180-\pi}$ and I will round the result.

Using Microsoft Excel, I found that $n=176$

I do not know if my approach is right. If right, how to solve for $u$. If wrong, how to solve this problem:

What is the least positive integer $n$ for which
  $|\sin(n)-\sin(n^{\circ})|<0.005$?

 A: Let $t:=\dfrac{\pi}{180}$ and $\epsilon:=0.005$.  We want to find the smallest integer $n>0$ such that $\big|\sin(n)-\sin(nt)\big|<\epsilon$.  That is,
$$2\,\Biggl|\sin\left(\frac{n(1-t)}{2}\right)\Biggr|\,\Biggl|\cos\left(\frac{n(1+t)}{2}\right)\Biggr|<\epsilon\,.$$
This shows that either 
$$\Biggl|\sin\left(\frac{n(1-t)}{2}\right)\Biggr|<\delta\text{ or }\Biggl|\cos\left(\frac{n(1+t)}{2}\right)\Biggr|<\delta\,,\tag{*}$$
where $\delta:=\sqrt{\dfrac{\epsilon}{2}}=0.05$.  Hence, there exists an integer $k>0$ such that
$$\left|\frac{n\left(1-(-1)^kt\right)}{2}-\frac{k\pi}{2}\right|<\text{ArcSin}(\delta)\,.$$
Let $d:=2\,\text{ArcSin}(\delta)\approx2\delta=0.01$.  Then,
$$n\in I_k:=\left(\frac{k\pi- d}{\left(1-(-1)^k\,t\right)},\frac{k\pi+d}{\left(1-(-1)^k\,t\right)}\right)\,.$$
Therefore, you have to only look for integers $k>0$ such that $I_k$ contains an integer $n_k$, and check whether the requirement $\big|\sin(n)-\sin(nt)\big|<\epsilon$ is satisfied when $n=n_k$.  Using a computer search from $k=1,2,\ldots$, we get that $k=57$ yields our smallest $n$, which is $176$.  The search is not too bad.  For the majority of cases, $I_k$ does not contain an integer.
However, you can also make a guess using the continued fractions of $\dfrac{\pi}{1-t}$ and $\dfrac{\pi}{1+t}$.  The continued fractions are, respectively,
$$\dfrac{\pi}{1-t}=3+\frac{1}{5+\frac{1}{15+\frac{1}{1+\ddots}}}\text{ and }\dfrac{\pi}{1+t}=3+\frac{1}{11+\frac{1}{2+\frac{1}{2+\ddots}}}\,.$$
The sequences of convergents are, respectively, $$\left\{3,\frac{16}{5},\frac{243}{76},\frac{259}{81},\ldots\right\}\text{ and }\left\{3,\frac{34}{11},\frac{71}{23},\frac{176}{57},\ldots\right\}\,.$$
The value of $\dfrac{n}{k}$ should be one of the convergents above.  The smallest $n$ which works and can be obtained by this method comes from the fraction $\dfrac{176}{57}$, giving the smallest value of $n$ to be $176$.  The reason that this method works is due to the fact that, for a real number $r$, if $p$ and $q$ are integers such that $q>0$, $\gcd(p,q)=1$, and $\dfrac{p}{q}$ is a convergent of $r$, then $\dfrac{p}{q}$ is a better estimate of $r$ than any rational number of the form $\dfrac{a}{b}$ with $a,b\in\mathbb{Z}$ such that $0<b\leq q$, i.e.,
$$\left|r-\dfrac{p}{q}\right|\leq \left|r-\dfrac{a}{b}\right|\,,$$
with the equality case $(a,b)=(p,q)$.  You even have a stronger inequality
$$|rq-p|\leq |rb-a|$$
where the equality holds iff $(a,b)=(p,q)$.
Interestingly, it is possible to find a positive integer $n$ such that both inequalities in (*) hold.  (For $n=176$, for example, only the right inequality with cosine is true.) The smallest $n$ that this happens is $n=2430$ (here is how I searched for it), where
$$\left|\sin\left(\frac{n(1-t)}{2}\right)\right|\approx 0.010959\approx\left|\cos\left(\frac{n(1+t)}{2}\right)\right|\,.$$
In fact,
$$\sin\left(\frac{n(1-t)}{2}\right)-\cos\left(\frac{n(1+t)}{2}\right)\approx 8.869\times 10^{-14}$$
for $n=2430$.  I am not sure what is so special about $n=2430$ as both  the sine and cosine values are so close to one another.  Especially, the next value of $n$ that makes both inequalities in (*) hold does not do the same thing, i.e., when $n=7111$, 
$$\left|\sin\left(\frac{n(1-t)}{2}\right)\right|\approx 0.00621\text{ but }\left|\cos\left(\frac{n(1+t)}{2}\right)\right|\approx 0.01124\,.$$
