What is $\sin(2)$ equal to? I understand that $\sin(2)$ is equal to $0.90929\ldots$. However, I am wondering if there is a simpler way of calculating this. Essentially I am wondering if there is an answer with this where $\pi$ is in there as an answer like $2\pi/6$ or something where it equals $0.90929\ldots$. 
 A: The sine function is transcendental, which means that generally, you cannot find a "simple" expression for $\sin(x)$. Nevertheless, you can approximate $\sin(x)$ (to which ever degree of precision you like) using Taylor polynomials.
A: The usual values of the sine and cosine functions are not expressions involving $\pi$ (the most familiar ones are $1,\frac12,\frac{\sqrt2}2,\frac{\sqrt 3}2$). You probably confuse with the arguments being rational multiples of $\pi$, i.e. fractions of a full turn.
In fact, the sine and cosine of $q\pi$ are always algebraic numbers, as $\sin nx, \cos nx$ can be developed as polynomials of degree $n$ in $\sin x,\cos x$. It is also known that for $q\pi$ the only rational values of the sine are $0,\pm1$ and $\pm\frac12$. (Check Niven's Theorem).
For arguments that are not in this form, I am not sure that much is known. As coined by Mohammad Zuhair Khan, $\sin 2$ is a transcendental number. For the same reason as before (polynomial transformation), $\sin n$ and $\cos n$ where $n$ is a natural must be transcendental numbers.
I don't think that there is another way to express $\sin 2$ using a closed-form expression (but trivially, $\Im e^{2i}$). Numerical evaluation is not a big deal with the Taylor expansions
$$\cos x=\sum_{k=0}^\infty\frac{(-x^2)^{k+1}}{(2k)!}$$
$$\sin x=x\sum_{k=0}^\infty\frac{(-x^2)^{k+1}}{(2k+1)!}.$$
For fast convergence, it is advisable to evaluate for a small power of $2$, say $2^{-8}$, then obtain the value for $2$ by successive doublings.
