To find the Taylor series for $f(x) = \sin(x)$ centered at $a = \pi$, we use the formula for the Taylor series:

$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$

First, we calculate the derivatives of $f(x) = \sin(x)$:

1. $f(x) = \sin(x)$
2. $f'(x) = \cos(x)$
3. $f''(x) = -\sin(x)$
4. $f'''(x) = -\cos(x)$
5. $f^{(4)}(x) = \sin(x)$ (and the cycle repeats)

Next, evaluate these at $x = \pi$:

• $f(\pi) = \sin(\pi) = 0$
• $f'(\pi) = \cos(\pi) = -1$
• $f''(\pi) = -\sin(\pi) = 0$
• $f'''(\pi) = -\cos(\pi) = 1$
• $f^{(4)}(\pi) = \sin(\pi) = 0$

Now substitute into the Taylor series:

$f(x) = 0 + (-1)(x - \pi) + \frac{0}{2!}(x - \pi)^2 + \frac{1}{3!}(x - \pi)^3 + 0 + \cdots$

Thus, the Taylor series simplifies to:

$f(x) = - (x - \pi) + \frac{1}{6}(x - \pi)^3 + \cdots$

The general form for the series, accounting for the periodicity of sine, is:

$\sin(x) = - (x - \pi) + \frac{1}{6}(x - \pi)^3 - \frac{1}{120}(x - \pi)^5 + \cdots$

Summary: The Taylor series for $\sin(x)$ centered at $\pi$ is:

$\sin(x) = - (x - \pi) + \frac{(x - \pi)^3}{6} - \frac{(x - \pi)^5}{120} + \cdots$

Would you like more details on any part of this derivation, or do you have any specific questions? Here are some related questions to consider:

1. How does the Taylor series converge for $\sin(x)$?
2. Can you derive the Taylor series for $\cos(x)$ centered at $\pi$?
3. What is the radius of convergence for the Taylor series of $\sin(x)$?
4. How does the Taylor series for $\sin(x)$ compare to its Maclaurin series?
5. Can you apply this series to approximate $\sin(\pi + x)$ for small $x$?

Tip: When working with Taylor series, always check the derivatives at the center point for a clear understanding of the series' behavior.