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$$.
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Math Problem Analysis
Mathematical Concepts
Calculus Taylor Series Trigonometry
Formulas
Taylor series expansion: f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3 + ... Derivatives of sin(x): f'(x) = cos(x), f''(x) = -sin(x), f'''(x) = -cos(x), ...