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To solve this problem, we want to convert the given sum into a product using a trigonometric identity.

The expression is:

\[
6 \left( \cos(\sqrt{5}x) + \cos(3\sqrt{5}x) \right)
\]

We can use the trigonometric identity for the sum of cosines:

\[
\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)
\]

Here, let \( A = \sqrt{5}x \) and \( B = 3\sqrt{5}x \).

Now, let's compute \( \frac{A + B}{2} \) and \( \frac{A - B}{2} \):

\[
\frac{A + B}{2} = \frac{\sqrt{5}x + 3\sqrt{5}x}{2} = \frac{4\sqrt{5}x}{2} = 2\sqrt{5}x
\]

\[
\frac{A - B}{2} = \frac{\sqrt{5}x - 3\sqrt{5}x}{2} = \frac{-2\sqrt{5}x}{2} = -\sqrt{5}x
\]

Substitute these back into the identity:

\[
\cos(\sqrt{5}x) + \cos(3\sqrt{5}x) = 2 \cos(2\sqrt{5}x) \cos(-\sqrt{5}x)
\]

Since \( \cos(-x) = \cos(x) \), we have:

\[
\cos(\sqrt{5}x) + \cos(3\sqrt{5}x) = 2 \cos(2\sqrt{5}x) \cos(\sqrt{5}x)
\]

Now, multiply by 6:

\[
6 \left( \cos(\sqrt{5}x) + \cos(3\sqrt{5}x) \right) = 6 \cdot 2 \cos(2\sqrt{5}x) \cos(\sqrt{5}x) = 12 \cos(2\sqrt{5}x) \cos(\sqrt{5}x)
\]

This matches the third option:

\[
\boxed{12 \cos(2\sqrt{5}x) \cos(\sqrt{5}x)}
\]

Would you like further details or have any questions?

### Related Questions to Explore:

1. How do you derive the trigonometric identities for sums to products?
2. What are other useful trigonometric identities for converting sums into products?
3. How can the trigonometric identity \( \cos A + \cos B \) be used in integration problems?
4. What is the significance of the identity \( \cos(-x) = \cos(x) \)?
5. How can the product-to-sum formulas be applied in physics?

#### Tip: 
Remember that trigonometric identities can simplify complex expressions and solve equations more efficiently!

Convert the sum 6[cos(√5x) + cos(3√5x)] to a product.

Explore the step-by-step solution for converting the sum 6[cos(√5x) + cos(3√5x)] into a product using trigonometric identities. This problem highlights the use of sum-to-product identities, making it suitable for students in Grades 10-12.

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