The Maclaurin series of \(f(x) = \sin(x)\) is:
\[ \sum^\infty_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \]
Let \(R_n(x)\) ne the \(n\)th remainder. According to the Lagrange form of the Taylor remainder, there exists an integer \(c\) such that:
\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \]
Let \(n\) go to infinity:
\[ \lim_{n \to \infty} R_n(x) = \lim_{n \to \infty} \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \]
Since \(f(x) = \sin(x)\), then all derivatives of \(f(x)\) would be one of: \(\sin(x)\), \(-\sin(x)\), \(\cos(x)\) or \(-\cos(x)\). This means \(f^{(n+1)}(c)\) will always be between 0 and 1 (inclusive). Also, as \(n\) goes to infinity, \((n+1)!\) will eventually grow faster than \(x^{n+1}\). Which means:
\[\begin{align} \lim_{n \to \infty} R_n(x) &= \lim_{n \to \infty} \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \\ &= 0 \end{align}\]
This shows that as \(n\) increases, the Maclaurin series of \(\sin(x)\) will approach \(\sin(x)\). A similar argument can be made for \(\cos(x)\), since the Maclaurin series for \(\cos(x)\) is:
\[ \sum^\infty_{n=0} (-1)^n \frac{x^{2n}}{(2n)!} \]