To find the Taylor series expansion of the function \(\sin(x)\) centered at \(x = \frac{\pi}{4}\), we will express \(\sin(x)\) as a Taylor series around \(x = \frac{\pi}{4}\).

### Taylor Series Formula

The Taylor series of a function \(f(x)\) centered at \(x = a\) is given by:

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

In this case, \(f(x) = \sin(x)\) and \(a = \frac{\pi}{4}\).

### Step-by-Step Calculation

1. **Compute the derivatives of \( \sin(x) \):**

\[
\begin{align*}
f(x) &= \sin(x), \\
f'(x) &= \cos(x), \\
f''(x) &= -\sin(x), \\
f'''(x) &= -\cos(x), \\
f^{(4)}(x) &= \sin(x), \quad \text{(and repeats in cycles)}
\end{align*}
\]

2. **Evaluate the derivatives at \( x = \frac{\pi}{4} \):**

\[
\begin{align*}
f\left(\frac{\pi}{4}\right) &= \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \\
f'\left(\frac{\pi}{4}\right) &= \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \\
f''\left(\frac{\pi}{4}\right) &= -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \\
f'''\left(\frac{\pi}{4}\right) &= -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \\
f^{(4)}\left(\frac{\pi}{4}\right) &= \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.
\end{align*}
\]

3. **Construct the Taylor series:**

Substitute these values into the Taylor series formula:

\[
\sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} \cdot (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12} \cdot (x - \frac{\pi}{4})^3 + \cdots
\]

### Simplified Taylor Series for \(\sin(x)\) around \(x = \frac{\pi}{4}\):

\[
\sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12} (x - \frac{\pi}{4})^3 + \cdots
\]

This gives the Taylor expansion up to the third degree. You can continue the series by calculating higher-order derivatives and including more terms as needed.

Would you like to see more details or have any questions about this process?

### Additional Related Questions:

1. How do you derive the Taylor series for other trigonometric functions like \(\cos(x)\) around different points?
2. What is the radius of convergence for the Taylor series of \(\sin(x)\)?
3. How can you use Taylor series to approximate \(\sin(x)\) for small values of \(x\)?
4. What is the significance of choosing a different center point for a Taylor series expansion?
5. How do you determine the error or remainder of a Taylor series approximation?

#### Tip:
When approximating a function using a Taylor series, the more terms you include, the closer the approximation will be to the original function, especially near the center point \(x = a\).

Explore the Taylor series expansion of sin(x) centered at x = π/4. This problem covers essential calculus concepts, including the Taylor series formula, trigonometric functions, and the computation of derivatives. Ideal for undergraduate calculus students.

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