Math Snap

STEP 1

Assumptions
1. The function given is $f(x) = \sin(x) \cos(x)$.
2. We need to find the first three terms of the Maclaurin series for this function.
3. The Maclaurin series is a Taylor series expansion of a function about $x = 0$.
4. The Maclaurin series for a function $f(x)$ is given by $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$.
5. We will calculate the derivatives of $f(x)$ and evaluate them at $x = 0$.

STEP 2

First, express $f(x) = \sin(x) \cos(x)$ using the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$.
$$f(x) = \frac{1}{2} \sin(2x)$$

STEP 3

Find the first derivative of $f(x)$.
$$f'(x) = \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x)$$

STEP 4

Find the second derivative of $f(x)$.
$$f''(x) = -2 \sin(2x)$$

STEP 5

Find the third derivative of $f(x)$.
$$f'''(x) = -4 \cos(2x)$$

STEP 6

Evaluate $f(x)$, $f'(x)$, $f''(x)$, and $f'''(x)$ at $x = 0$.
1. $f(0) = \frac{1}{2} \sin(0) = 0$
2. $f'(0) = \cos(0) = 1$
3. $f''(0) = -2 \sin(0) = 0$
4. $f'''(0) = -4 \cos(0) = -4$

STEP 7

Substitute these values into the Maclaurin series formula.
$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

STEP 8

Calculate the first three terms of the Maclaurin series.
1. $f(0) = 0$
2. $f'(0)x = 1 \cdot x = x$
3. $\frac{f''(0)}{2!}x^2 = \frac{0}{2}x^2 = 0$
4. $\frac{f'''(0)}{3!}x^3 = \frac{-4}{6}x^3 = -\frac{2}{3}x^3$

STEP 9

Combine the terms to get the first three non-zero terms of the Maclaurin series.
$$f(x) = x - \frac{2}{3}x^3$$

Compare the calculated series with the given options.
The series $x - \frac{2}{3}x^3$ matches the first two terms of option A: $x - \frac{2 x^{3}}{3} + \frac{2 x^{4}}{15}$.
Therefore, the correct answer is option A.