91Ó°ÊÓ

We need to find the general expression for the nth derivative of \(f(x) = \sin x\). We can do this by taking the first few derivatives and identifying the pattern. $$f'(x) = \cos x$$ $$f''(x) = -\sin x$$ $$f^{(3)}(x) = -\cos x$$ $$f^{(4)}(x) = \sin x$$ Now we can see the pattern: the derivatives of \(\sin x\) repeat every four derivatives. Therefore, we have: $$f^{(n)}(x) = \begin{cases} \sin x & \text{if } n \equiv 0 \pmod{4} \\ \cos x & \text{if } n \equiv 1 \pmod{4} \\ -\sin x & \text{if } n \equiv 2 \pmod{4} \\ -\cos x & \text{if } n \equiv 3 \pmod{4} \end{cases}$$

Since the Taylor polynomial is centered at \(a = 0\), substitute \(a = 0\) into \(f^{(n)}(x)\). $$f^{(n)}(0) = \begin{cases} \sin 0 = 0 & \text{if } n \equiv 0 \pmod{4} \\ \cos 0 = 1 & \text{if } n \equiv 1 \pmod{4} \\ -\sin 0 = 0 & \text{if } n \equiv 2 \pmod{4} \\ -\cos 0 = -1 & \text{if } n \equiv 3 \pmod{4} \end{cases}$$

Substitute the expression for \(f^{(n)}(0)\) into the Lagrange remainder formula and express the result for a general value of \(n\). $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-0)^{n+1} = \frac{x^{n+1}}{(n+1)!}f^{(n+1)}(c)$$ Using the expression for \(f^{(n)}(x)\), the remainder can be written as: $$R_n(x) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod{4} \\ \frac{x^{n+1}}{(n+1)!}\cos c & \text{if } n \equiv 1 \pmod{4} \\ 0 & \text{if } n \equiv 2 \pmod{4} \\ -\frac{x^{n+1}}{(n+1)!}\cos c & \text{if } n \equiv 3 \pmod{4} \end{cases}$$ So, the remainder of the nth-order Taylor polynomial centered at \(a = 0\) for the function \(f(x) = \sin x\) is: $$R_n(x) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod{4} \\ \frac{x^{n+1}}{(n+1)!}\cos c & \text{if } n \equiv 1 \pmod{4} \\ 0 & \text{if } n \equiv 2 \pmod{4} \\ -\frac{x^{n+1}}{(n+1)!}\cos c & \text{if } n \equiv 3 \pmod{4} \end{cases}$$