To find the third-order Maclaurin polynomial for $f(x) = \sin(4x)\cos(x)$, we will expand $f(x)$ around $x = 0$ using Taylor series.

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Step 1: Rewrite $f(x)$

Using the product-to-sum identity: $\sin(4x)\cos(x) = \frac{1}{2} \left[ \sin(5x) + \sin(3x) \right]$ Thus, $f(x) = \frac{1}{2} \left[ \sin(5x) + \sin(3x) \right].$

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Step 2: Expand $\sin(5x)$ and $\sin(3x)$ as Taylor series

The Taylor series expansion for $\sin(kx)$ around $x = 0$ is: $\sin(kx) = kx - \frac{(kx)^3}{6} + \mathcal{O}(x^5).$

For $\sin(5x)$: $\sin(5x) = 5x - \frac{(5x)^3}{6} = 5x - \frac{125x^3}{6}.$

For $\sin(3x)$: $\sin(3x) = 3x - \frac{(3x)^3}{6} = 3x - \frac{27x^3}{6}.$

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Step 3: Combine the series

Now substitute these expansions into $f(x)$: $f(x) = \frac{1}{2} \left[ \sin(5x) + \sin(3x) \right].$ $f(x) = \frac{1}{2} \left[ (5x - \frac{125x^3}{6}) + (3x - \frac{27x^3}{6}) \right].$

Simplify: $f(x) = \frac{1}{2} \left[ 8x - \frac{125x^3}{6} - \frac{27x^3}{6} \right].$ $f(x) = \frac{1}{2} \left[ 8x - \frac{152x^3}{6} \right].$ $f(x) = 4x - \frac{76x^3}{6}.$ $f(x) = 4x - \frac{38x^3}{3}.$

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Step 4: The third-order Maclaurin polynomial

The third-order Maclaurin polynomial of $f(x)$ is: $P_3(x) = 4x - \frac{38x^3}{3}.$

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Would you like further details or have any questions?

Related Questions:

1. How do you find the Maclaurin polynomial for $\cos(x)$ instead of $\sin(x)$?
2. What is the general process for Taylor series expansions for trigonometric functions?
3. How would the fourth-order polynomial differ from the third-order polynomial for this function?
4. Can you use differentiation to verify this polynomial?
5. What happens if we approximate $\sin(4x)$ and $\cos(x)$ individually and then multiply?

Tip:

For product functions, simplifying with identities before expanding often reduces complexity.