To find the Maclaurin series of \( \cos(3x) \), we need to expand \( \cos(3x) \) around \( x = 0 \). The Maclaurin series for a function \( f(x) \) is:

\[
f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots
\]

### Step-by-Step Solution

1. **Identify \( f(x) = \cos(3x) \) and compute derivatives at \( x = 0 \):**

   We know:
   \[
   f(x) = \cos(3x)
   \]

2. **Compute the derivatives and evaluate them at \( x = 0 \):**

   - \( f(x) = \cos(3x) \), so \( f(0) = \cos(0) = 1 \).
   
   - First derivative: \( f'(x) = -3 \sin(3x) \).  
     At \( x = 0 \): \( f'(0) = -3 \sin(0) = 0 \).
   
   - Second derivative: \( f''(x) = -9 \cos(3x) \).
     At \( x = 0 \): \( f''(0) = -9 \cos(0) = -9 \).
   
   - Third derivative: \( f'''(x) = 27 \sin(3x) \).
     At \( x = 0 \): \( f'''(0) = 27 \sin(0) = 0 \).
   
   - Fourth derivative: \( f^{(4)}(x) = -81 \cos(3x) \).
     At \( x = 0 \): \( f^{(4)}(0) = -81 \cos(0) = -81 \).

3. **Construct the series:**

   Now substitute these values into the Maclaurin series formula:

   \[
   f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \cdots
   \]

   Substitute the values we calculated:

   \[
   \cos(3x) = 1 + 0 \cdot x + \frac{-9}{2} x^2 + 0 \cdot x^3 + \frac{-81}{24} x^4 + \cdots
   \]

4. **Simplify terms:**

   This simplifies to:

   \[
   \cos(3x) = 1 - \frac{9}{2} x^2 - \frac{81}{24} x^4 + \cdots
   \]

   Further simplifying \( \frac{81}{24} = \frac{27}{8} \), we get:

   \[
   \cos(3x) = 1 - \frac{9}{2} x^2 + \frac{27}{8} x^4 + \cdots
   \]

### Final Answer

The Maclaurin series for \( \cos(3x) \) is:

\[
\cos(3x) = 1 - \frac{9}{2} x^2 + \frac{27}{8} x^4 + \cdots
\]

Would you like more details on each step or have any questions?

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### Related Questions
1. What is the Maclaurin series of \( \sin(3x) \)?
2. How does the Maclaurin series for \( \cos(x) \) compare to \( \cos(3x) \)?
3. What are the convergence properties of the Maclaurin series for cosine functions?
4. How would you find the Maclaurin series of \( e^{3x} \)?
5. How can the Taylor series expansion of \( \cos(x) \) help in calculating higher-order derivatives?

### Tip:
The Maclaurin series for trigonometric functions like \( \sin(x) \) and \( \cos(x) \) always have alternating signs and factorial denominators, which is a helpful pattern to remember.

Explore the step-by-step process to find the Maclaurin series for cos(3x), covering derivative calculations and power series expansion. This detailed solution is ideal for students studying calculus, focusing on series expansions and trigonometric functions.

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