How do I compute an infinite sine (or cosine) Taylor expansion? 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
 A: 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.

  
*
  
*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.


  
*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.
