how to compare $\sin(19^{2013}) $ and $\cos(19^{2013})$

Question

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!

What's the source of the problem? I don't think the abstract algebra tag fits. — Tyler, Jul 03 '13 at 02:40
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 03 '13 at 02:44
$19^{2013}$ looks like part of the date the assignment was given (June 19th, 2013). :P Anyways, correct me if I'm wrong, but the answer would be the same as comparing $\sin(1)$ and $\cos(1)$, because it's relative. Right? Looks a bit like a trick question to me... — Elliot Bonneville, Jul 03 '13 at 02:56
@StevenStadnicki,the range means such as $\frac{\sqrt{2}}{2}<sinA<\frac{\sqrt{3}}{2}.$ — chenbai, Jul 03 '13 at 03:00
@ElliotBonneville, ya, at begin I have same feeling but I can't go further. — chenbai, Jul 03 '13 at 03:03
it is trivial that $\frac{\pi}{2}>1>\frac{\pi}{4} \to sin(1)>cos(1)$ — chenbai, Jul 03 '13 at 03:06
I guess you could be tricky and find $19^{2013} \mod 360$ and go from there? — Tyler, Jul 03 '13 at 03:08
There's your answer then. Since 1 = 1 and $19^{2013}$ = $19^{2013}$, $sin(19^{2013}) > cos(19^{2013})$. — Elliot Bonneville, Jul 03 '13 at 03:08
@ElliotBonneville,what is the reason? $sin(19^3)< cos(19^3)$ — chenbai, Jul 03 '13 at 03:14
Overly simplistic approach due to lack of knowledge led to plausibly incorrect conclusion on my part. Apologies. — Elliot Bonneville, Jul 03 '13 at 03:17
@Tyler,it is radian, BTW, it is a high school exec and mod is not taught. — chenbai, Jul 03 '13 at 03:19
@ElliotBonneville,that is OK. I went first with your result also but then I found the problem and post it. thanks for your kindness. — chenbai, Jul 03 '13 at 03:29
Is the angle intended to be in radians or degrees? If it is in degrees modding by 360 solves it easily. — Spencer, Jul 03 '13 at 03:55
@Spencer,it is radians. But I am also interesting to see the solution in degrees as Tyler suggested also. — chenbai, Jul 03 '13 at 03:59
@LordSoth,Ha, $19^{2013}=19^{2012}\times 19=(360n+1) \times 19= 360k+19$, thanks! — chenbai, Jul 03 '13 at 04:10

Spencer · Answer 1 · 2013-07-03T04:29:33.967

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 $360$.

thanks ,Spencer.(+1).Now I am waiting the solution for radians. — chenbai, Jul 03 '13 at 04:25
No problem :). I'm not sure how to solve it in raidans yet, but I will try to figure it out. — Spencer, Jul 03 '13 at 04:30
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 03 '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 04 '13 at 01:31

score 8 · Accepted Answer · edited Apr 13 '17 at 12:19

8

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.

edited Apr 13 '17 at 12:19

Community

1

answered Jul 03 '13 at 05:24

Dan

14,978

1

I'd say that's cheating :) – Nathaniel Bubis Jul 03 '13 at 06:18
1

Overkill much? :p – Thomas Jul 03 '13 at 06:25
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 03 '13 at 06:53

how to compare $\sin(19^{2013}) $ and $\cos(19^{2013})$

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