1. **State the problem:** We want to find the Taylor series expansion of the function $f(x) = \cos x$ centered at $x=0$. 2. **Recall the Taylor series formula:** The Taylor series of a function $f(x)$ at $x=a$ is given by $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x - a)^n$$ where $f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at $x=a$. 3. **Calculate derivatives of $\cos x$ and evaluate at $x=0$:** - $f(x) = \cos x$, so $f(0) = 1$ - $f'(x) = -\sin x$, so $f'(0) = 0$ - $f''(x) = -\cos x$, so $f''(0) = -1$ - $f^{(3)}(x) = \sin x$, so $f^{(3)}(0) = 0$ - $f^{(4)}(x) = \cos x$, so $f^{(4)}(0) = 1$ 4. **Write the Taylor series using the nonzero terms:** Only even derivatives are nonzero at 0, so the series is $$\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ 5. **Explanation:** The Taylor series for $\cos x$ at 0 consists of only even powers of $x$ with alternating signs because the odd derivatives of $\cos x$ at 0 are zero. 6. **Final answer:** $$\boxed{\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}}$$