To solve this problem, we will find the Taylor Series expansion of $f(x) = \sin x$ about $a = \frac{\pi}{4}$.

The Taylor Series for a function $f(x)$ about $a$ is given by: $f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$

Step 1: Compute the derivatives of $\sin x$

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

Step 2: Evaluate the derivatives at $x = \frac{\pi}{4}$

At $x = \frac{\pi}{4}$:

1. $f(a) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
2. $f'(a) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
3. $f''(a) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
4. $f'''(a) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

Step 3: Construct the Taylor Series

Substitute into the Taylor Series formula: $f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$ $f(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \frac{-\frac{\sqrt{2}}{2}}{2!}(x - \frac{\pi}{4})^2 + \frac{-\frac{\sqrt{2}}{2}}{3!}(x - \frac{\pi}{4})^3 + \cdots$

Simplify the terms: $f(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 + \cdots$

This is the Taylor Series expansion for $\sin x$ about $a = \frac{\pi}{4}$.

Would you like further explanation or additional examples?

Related Questions:

1. How is the Taylor Series formula derived?
2. How does the radius of convergence apply to Taylor Series expansions?
3. What is the importance of higher-order terms in Taylor Series approximations?
4. Can Taylor Series expansions be used for functions other than trigonometric ones?
5. What happens if you expand around a different point, such as $a = 0$?

Tip:

To simplify calculations, recognize periodic patterns in derivatives, especially for trigonometric functions!