find the taylor series expansion of the function cosx  aboutx  x=pi/4

To find the Taylor series expansion of the function \( \cos(x) \) about \( x = \frac{\pi}{4} \), we need to determine the function's derivatives at \( x = \frac{\pi}{4} \) and then construct the series.

The Taylor series expansion of \( \cos(x) \) about \( x = \frac{\pi}{4} \) is given by:

\[
\cos(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}\left(\frac{\pi}{4}\right)}{n!} \left(x - \frac{\pi}{4}\right)^n
\]

where \( f^{(n)}\left(\frac{\pi}{4}\right) \) represents the \( n \)-th derivative of \( \cos(x) \) evaluated at \( x = \frac{\pi}{4} \).

### Step 1: Compute the derivatives of \( \cos(x) \)
Let's compute the first few derivatives of \( \cos(x) \):

\[
f(x) = \cos(x)
\]

\[
f'(x) = -\sin(x)
\]

\[
f''(x) = -\cos(x)
\]

\[
f'''(x) = \sin(x)
\]

\[
f^{(4)}(x) = \cos(x)
\]

The pattern repeats every four derivatives.

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

\[
f\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]

\[
f'\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}
\]

\[
f''\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}
\]

\[
f'''\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]

### Step 3: Write the Taylor series

Substituting these into the Taylor series formula, we have:

\[
\cos(x) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \cdot \left(x - \frac{\pi}{4}\right) - \frac{\sqrt{2}}{4} \cdot \left(x - \frac{\pi}{4}\right)^2 + \frac{\sqrt{2}}{12} \cdot \left(x - \frac{\pi}{4}\right)^3 + \cdots
\]

### Final Taylor Series Expansion

The Taylor series expansion of \( \cos(x) \) about \( x = \frac{\pi}{4} \) is:

\[
\cos(x) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\left(x - \frac{\pi}{4}\right) - \frac{\sqrt{2}}{4}\left(x - \frac{\pi}{4}\right)^2 + \frac{\sqrt{2}}{12}\left(x - \frac{\pi}{4}\right)^3 + \cdots
\]

This is the Taylor series expansion of \( \cos(x) \) centered at \( x = \frac{\pi}{4} \).

Would you like more details, or do you have any questions?

### Related Questions:
1. How do you find the radius of convergence for the Taylor series?
2. Can you find the Taylor series of \( \cos(x) \) about \( x = 0 \) (Maclaurin series)?
3. What are the applications of Taylor series in physics?
4. How can you approximate \( \cos(x) \) near \( x = \frac{\pi}{4} \) using the first few terms of the series?
5. How does the Taylor series change if the function was \( \sin(x) \) instead of \( \cos(x) \)?

### Tip:
To improve accuracy, the more terms you include in the Taylor series, the better the approximation of the function near the point of expansion.

find the Taylor series expansion of the function cos(x) about x = pi/4

Explore the Taylor series expansion of cos(x) around x = π/4. This problem involves calculus concepts, including the derivatives of trigonometric functions and the application of Taylor's Theorem. Suitable for students in Grades 11-12 or college-level calculus courses.

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