<s>[INST] 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 $0.5$
 [/INST] 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.
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