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.