Find the first four terms of the Taylor series for the function 3cos(x) about the point a=−π/4. (Your answers should include the variable x when appropriate.)

degree 0 term =
degree 1 term =
degree 2 term =
degree 3 term =

To find the first four terms of the Taylor series for the function \( f(x) = 3 \cos(x) \) about the point \( a = -\frac{\pi}{4} \), we need to compute the Taylor series expansion using the formula:

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

Let's compute the derivatives and evaluate them at \( a = -\frac{\pi}{4} \):

### 1. Function and Derivatives
- \( f(x) = 3 \cos(x) \)
- First derivative: \( f'(x) = -3 \sin(x) \)
- Second derivative: \( f''(x) = -3 \cos(x) \)
- Third derivative: \( f'''(x) = 3 \sin(x) \)

### 2. Evaluate at \( a = -\frac{\pi}{4} \):
- \( f\left(-\frac{\pi}{4}\right) = 3 \cos\left(-\frac{\pi}{4}\right) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \)
- \( f'\left(-\frac{\pi}{4}\right) = -3 \sin\left(-\frac{\pi}{4}\right) = -3 \cdot \left(-\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} \)
- \( f''\left(-\frac{\pi}{4}\right) = -3 \cos\left(-\frac{\pi}{4}\right) = -\frac{3\sqrt{2}}{2} \)
- \( f'''\left(-\frac{\pi}{4}\right) = 3 \sin\left(-\frac{\pi}{4}\right) = 3 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2} \)

### 3. Taylor Series Expansion:
Now, substituting the function values and derivatives into the Taylor series formula:

- Degree 0 term (constant term):  
  \( f(a) = \frac{3\sqrt{2}}{2} \)

- Degree 1 term (linear term):  
  \( f'(a)(x - a) = \frac{3\sqrt{2}}{2}(x + \frac{\pi}{4}) \)

- Degree 2 term (quadratic term):  
  \( \frac{f''(a)}{2!}(x - a)^2 = \frac{-\frac{3\sqrt{2}}{2}}{2}(x + \frac{\pi}{4})^2 = -\frac{3\sqrt{2}}{4}(x + \frac{\pi}{4})^2 \)

- Degree 3 term (cubic term):  
  \( \frac{f'''(a)}{3!}(x - a)^3 = \frac{-\frac{3\sqrt{2}}{2}}{6}(x + \frac{\pi}{4})^3 = -\frac{3\sqrt{2}}{12}(x + \frac{\pi}{4})^3 = -\frac{\sqrt{2}}{4}(x + \frac{\pi}{4})^3 \)

### Final Answer:
- Degree 0 term = \( \frac{3\sqrt{2}}{2} \)
- Degree 1 term = \( \frac{3\sqrt{2}}{2}(x + \frac{\pi}{4}) \)
- Degree 2 term = \( -\frac{3\sqrt{2}}{4}(x + \frac{\pi}{4})^2 \)
- Degree 3 term = \( -\frac{\sqrt{2}}{4}(x + \frac{\pi}{4})^3 \)

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How do you find higher-order terms in a Taylor series?
2. What is the radius of convergence for this Taylor series?
3. How do you find the Taylor series for a function about a point \( a \)?
4. How does the Taylor series change if we expand around a different point \( a \)?
5. How do you use the Taylor series to approximate values of functions near a point?

**Tip:** The more terms you include in a Taylor series, the better it approximates the function near the point of expansion.

Find the first four terms of the Taylor series for the function 3cos(x) about the point a = −π/4.

Find the first four terms of the Taylor series for 3cos(x) expanded about the point a = -π/4. This solution includes detailed steps involving trigonometric derivatives and the Taylor series formula. Suitable for students learning undergraduate calculus.

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