To determine the Taylor expansion of $\cos \theta$ around $\theta = \pi$, we need to expand $\cos(\theta)$ at $\theta = \pi$ up to the fourth order.

The Taylor series expansion of a function $f(\theta)$ around a point $\theta = a$ is given by: $f(\theta) = f(a) + f'(a)(\theta - a) + \frac{f''(a)}{2!}(\theta - a)^2 + \frac{f'''(a)}{3!}(\theta - a)^3 + \frac{f^{(4)}(a)}{4!}(\theta - a)^4 + O((\theta - a)^5).$

For $f(\theta) = \cos(\theta)$ at $a = \pi$:

1. $f(\pi) = \cos(\pi) = -1$.
2. The first derivative $f'(\theta) = -\sin(\theta)$, so $f'(\pi) = -\sin(\pi) = 0$.
3. The second derivative $f''(\theta) = -\cos(\theta)$, so $f''(\pi) = -\cos(\pi) = 1$.
4. The third derivative $f'''(\theta) = \sin(\theta)$, so $f'''(\pi) = \sin(\pi) = 0$.
5. The fourth derivative $f^{(4)}(\theta) = \cos(\theta)$, so $f^{(4)}(\pi) = \cos(\pi) = -1$.

Substituting these values into the Taylor series, we get: $\cos(\theta) \approx -1 + \frac{1}{2}(\theta - \pi)^2 - \frac{1}{24}(\theta - \pi)^4 + O((\theta - \pi)^5).$

From the provided options, this matches:

• Option D: $-1 + \frac{1}{2}(\theta - \pi)^2 + O((\theta - \pi)^4)$.

Correct Answer: D

Would you like more details on Taylor expansions or have any questions?

Here are some additional questions to expand on this topic:

1. How do Taylor series approximations improve with more terms?
2. What are the practical applications of Taylor series in physics or engineering?
3. How does the radius of convergence affect a Taylor series expansion?
4. Can Taylor series be used for functions with singularities?
5. What are the differences between Taylor and Maclaurin series?

Tip: For functions centered at zero, the Taylor series is called the Maclaurin series.