Is $\cos(\frac{\pi}{3})$ exactly equal to 0.5 or approximately equal to 0.5 We know that $\cos(\frac{\pi}{3})=\frac{1}{2}$, but the expansion for $\cos(x)$ is as follows: 
$$ \cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$$
So this would make
$$\begin{align}
\cos(\frac{\pi}{3}) & = 1-\frac{(\frac{\pi}{3})^2}{2!}+\frac{(\frac{\pi}{3})^4}{4!}-\frac{(\frac{\pi}{3})^6}{6!}+\ldots\\
\Rightarrow\frac{1}{2} & = 1-\frac{(\frac{\pi}{3})^2}{2!}+\frac{(\frac{\pi}{3})^4}{4!}-\frac{(\frac{\pi}{3})^6}{6!}+\ldots\qquad{(1)} 
\end{align}$$
Here the $LHS$ is rational, but the $RHS$ appears to be irrational. So would it be correct to say that $\cos(\frac{\pi}{3})$ is approximately equal to $0.5$ or is there any proof that the $RHS$ in equation $(1)$ is exactly equal to $0.5$
 A: Yes there are proofs that $\cos\frac{\pi}{3}=\frac{1}{2}$ exactly. It's the real part of either primitive sixth root of unity closest to $1$. These are the roots of the cyclotomic polynomial $\Phi_6(x)=\frac{(x^6-1)(x-1)}{(x^3-1)(x^2-1)}=x^2-x+1$.
By Vieta's formulas the sum of these roots is $-(-1)=1$, so $\cos\frac{\pi}{3}$ is half of this.
Note that the rationality/irrationality of the terms in an infinite series have absolutely nothing to do with the rationality/irrationality of the resulting sum's value. For instance,
$$\pi=3.1415\cdots=3+\frac{1}{10}+\frac{4}{100}+\frac{1}{1000}+\frac{5}{10000}+\cdots$$
is a sum of rational numbers, but $\pi$ is irrational. Conversely,
$$(1-\pi^{-1})+(\pi^{-1}-\pi^{-2})+(\pi^{-2}+\pi^{-3})+\cdots=1$$
is a series of irrational terms whose sum's value is rational. As Nishant points out in the comments, even a finite sum of irrational numbers may be rational, e.g. $(1-\sqrt{2})+\sqrt{2}=1$.
A: Both sides are rational. Both sides are equal to $\frac12$. This Taylor series converges everywhere to $\cos x$.
Just because you put an irrational value in for $x$ certainly doesn't mean that you must get an irrational value out for the sum of the series. After all,
$$2=e^{\log 2} = 1+\log 2+(\log 2)^2/2!+(\log 2)^3/3! +\cdots$$
yet $\log 2$ is irrational.
Long story short: A limit of irrationals need not be irrational.
Another example: $\pi^{-n}\rightarrow 0$, and $0$ is rational, yet each of the terms in this sequence are irrational.
