The Taylor expansion of sine and cosine are given by:

$$\begin{align} \sin(x)&= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\\ \cos(x)&= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \end{align}$$

And so, given any radian, I would like to be able to calculate exactly the value of this sum. And also, I want a method that always gives me an rational number when the series converges to a rational number. That is, one that gives 1 when I plug $\pi/2$ into sine, rather than an arbitrarily close approximation

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    $\begingroup$ This question does not make sense. A Taylor series is necessarily an approximation. The way we get exact values is by summing the series to the relevant function and then using various methods on that function. $\endgroup$ – Ryan Unger Jan 15 '15 at 19:57
  • $\begingroup$ What I mean is, if I plug pi into the sine expansion, I get pi - pi^3/3! + pi^5/5! - pi^7/7! + ... , that is, an infinite series in with terms involving pi. This series sums up for a total of 1. $\endgroup$ – user207032 Jan 15 '15 at 20:45
  • $\begingroup$ If I were to plug values like pi/2, pi, 3pi/2, 2pi, and so on into the expansion, I would get infinite series which converge to rational numbers, just like geometric series do. So, when this happens, how do I do the summation? $\endgroup$ – user207032 Jan 15 '15 at 20:47
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    $\begingroup$ Of course. The reason you know that is because $\sin(\pi)=1$. The reason we know that $\pi-\pi^3/3+\cdots$ sums up to $1$ is because it is the sine function. There is no way to keep summing terms and get the exact answer. Infinite geometric series cannot be evaluated either by simply summing. You know the precise answer because it can be summed to $1/(1-x)$ or something. $\endgroup$ – Ryan Unger Jan 15 '15 at 20:49
  • $\begingroup$ However there are ways of circumvent so that we do not have to actually do an infinite sum, and this is what I am asking. $\endgroup$ – user207032 Jan 15 '15 at 21:12

Although the user's deleted their account, I'll provide an answer since there isn't one. Of course, this means I won't know if I misinterpreted their question.

  1. What is the exact value of the sum of the Taylor Series of $\sin(x)$ and $\cos(x)$?

Clearly, the values are $\sin(x)$ and $\cos(x)$, by definition. Whether this value is a particular rational or algebraic number is another question. If we could easily determine a simple, exact value for such functions, we wouldn't bother with a Taylor Series.

  1. Given $x$ and that $\sin(x)$ or $\cos(x)$ is rational, what possible methods can determine the value of $\sin(x)$ or $\cos(x)$?

Since both $\sin(x)$ and $\cos(x)$ can equal every every number in $[-1,1]$, they can take countably infinitely many rational values. It is known by Niven's Theorem that the only $x$ in $[0,90^\circ]$ such that both $x^\circ$ and $\sin(x^\circ)$ are rational are $x=0,30^\circ,90^\circ$.

It follows that if $x=\frac{\pi}{180}a$ for $x$ in $[0,\pi/2]$ then both $a$ and $\sin(x)$ can be rational only when $a=0,30,90$. We also know, by Henning Makholm's post re. the Lindemann-Weierstrass Theorem, that if $\sin(x)$ or $\cos(x)$ is rational then $x$ cannot be nonzero algebraic.

So the only nontrivial $x$ we can actually be given are categorically neither algebraic nor rational multiples of $\pi$. This makes what appeared to be a tricky problem into an incredibly difficult one. Even if I gave you an $x$ that fulfilled the criteria, it would likely be a long definition of a constant that we know next to nothing about. We don't even know whether relatively simple constants like $\pi+e$ are algebraic or not so we can't say whether their $\sin$ would be rational.

However, there might be some use in using the inverse trigonometric functions in terms of complex logarithms. As $\arccos(y)=-i\ln(y+\sqrt{y^2-1})$, we can see that if $\cos(x)=\frac{p}{q}$ is rational then if we were able to manipulate $x$ into the form $-i\ln\left(\frac{p}{q}+\sqrt{\frac{p^2}{q^2}-1}\right)$, we could find $\frac{p}{q}$.

For example, let's say we're given $x=i\ln\left(\frac{5}{1+2\sqrt{6}i}\right)\approx1.369$, then we can show that $x=-i\ln\left(\frac{1}{5}+\sqrt{\left(\frac{1}{5}\right)^2-1}\right)$ so $\cos(x)=\frac{1}{5}$. So it might be worth seeing if this question can be reduced to one about $i\ln(r)$, where $r$ is complex algebraic.

Regards, Jam.

If anyone could drop me a comment if they think I'm wrong about anything, please do so. Thanks.


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