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how to compare $ \sin(19^{2013})$ and $\cos (19^{2013})$ or even find their value range with normal calculator?

I can take $2\pi k= 19^{2013} \to \ln(k)= 2013 \ln(19)- \ln(2 \pi)=5925.32 \to k= 2.089 \times 10^{5925}$, but it useless.(I can get final answer with WolframAlpha but it is not allowed.)

Any hint? thanks!

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What's the source of the problem? I don't think the abstract algebra tag fits. – Tyler Jul 3 '13 at 2:40
How do you define 'value range'? Obviously sin and cos will be between -1 and 1, but to get any more information than that you'll need to know $\pi$ to roughly 2500 places or so... – Steven Stadnicki Jul 3 '13 at 2:44
@Tyler,it is radian, BTW, it is a high school exec and mod is not taught. – chenbai Jul 3 '13 at 3:19
Is the angle intended to be in radians or degrees? If it is in degrees modding by 360 solves it easily. – Spencer Jul 3 '13 at 3:55
$19^2 = 360+1$. – Lord Soth Jul 3 '13 at 4:04
up vote 8 down vote accepted

To find the sine and cosine, you'll need to reduce the angle ($19^{2013} \approx 1.352 \times 10^{2574}$) modulo $2\pi$. So, for reasonable accuracy, you'll need about 2600 digits of $\pi$. Fortunately, the first 100 000 or even million digits are readily accessible online.

It happens that $19^{2013} \approx 1.2329141525482654$ modulo $2\pi$, so:

  • $\sin(19^{2013}) \approx \sin(1.2329141525482654) \approx 0.9434588183383549$
  • $\cos(19^{2013}) \approx \cos(1.2329141525482654) \approx 0.3314897556480367$

All you need is a programming language that supports arbitrary-precision rational arithmetic, and a way of obtaining lots of digits of $\pi$. A normal TI-89 calculator will provide the former; the hard part is implementing an algorithm for the latter. But that's a topic for another question.

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I'd say that's cheating :) – nbubis Jul 3 '13 at 6:18
Overkill much? :p – Thomas Jul 3 '13 at 6:25
Woo, that is very interesting and nice answer.(+1) so it seems that I have to learn some programming if only with the calculator. I will wait another day to see some surprising answers to post. – chenbai Jul 3 '13 at 6:53

If $19^{2013}$ were measured in degrees we could use a simple trick.

$19$ is one more than $18$ which is a multiple of $6$. This means that any power of $19$ is also $1$ more than a multiple of $6$, in particular $19^2 = 361$

$19^2$ being $1$ more than a multiple of $360$ tells us that any even power of $19$ will be 1 more than a multiple of $360$. In particular we can conclude that $19^{2012}$ is one more than a multiple of $360$.

What does that mean? It means that if we travel $19^{2012}$ degrees around the unit circle we will just end up at the $1$ degree tick mark.

$19^{2013}$ is just $19^{2012}$ $19$ times. This means we would make the trip described above $19$ times each time ending up one tick further than the last time meaning that we would finish at $19$ degrees. In other words $19$ degrees is located at the same position of the unit circle as $19^{2013}$.

If you've never worked with modular arithmetic you may not understand why $19^2 = 360 + 1$ implies that $19^{2012} = k \cdot 360+1$.

To see this consider powers of $(360+1)$.

$$360+1$$ $$ (360+1)^2 = 360^2 + 2\cdot360 + 1$$ $$ (360+1)^3 = 360^3 + 3^360^2 + 3*360 + 1$$ $$ \vdots$$ $$ (360+1)^n = 360^n + n *360^{n-1} + \cdots + n * 360 + 1 $$

Notice the result is always one more than a multiple of $360$.

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thanks ,Spencer.(+1).Now I am waiting the solution for radians. – chenbai Jul 3 '13 at 4:25
No problem :). I'm not sure how to solve it in raidans yet, but I will try to figure it out. – Spencer Jul 3 '13 at 4:30
+1 nice solution – Ovi Jul 3 '13 at 7:49
There is no easy way to answer in radians; it is a variation of the famous "problem that stumped Feynman", to calculate $\tan 10^{100}$ to within ten percent. – MJD Jul 3 '13 at 17:17
@MJD,thanks, following your comments, I just learn the story of Feynman that some one post it in the answer of SE also. – chenbai Jul 4 '13 at 1:31

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