To express the product $\sin(3x) \cos(4x)$ as a sum, we use the product-to-sum identity:

$\sin A \cos B = \frac{1}{2} \left[ \sin(A+B) + \sin(A-B) \right]$

Substituting $A = 3x$ and $B = 4x$:

$\sin(3x) \cos(4x) = \frac{1}{2} \left[ \sin(3x + 4x) + \sin(3x - 4x) \right]$

$= \frac{1}{2} \left[ \sin(7x) + \sin(-x) \right]$

Since $\sin(-x) = -\sin(x)$, we get:

$\sin(3x) \cos(4x) = \frac{1}{2} \left[ \sin(7x) - \sin(x) \right]$

This is the sum representation of the given product.

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Would you like more details or have any questions?

Here are 5 related problems to explore further:

1. Express $\cos(5x) \sin(2x)$ as a sum.
2. Write $\cos(3x) \cos(4x)$ as a sum.
3. Convert $\sin(6x) \cos(2x)$ into a sum using product-to-sum identities.
4. Derive the product-to-sum identities from sum-to-product identities.
5. Use product-to-sum identities to simplify $\sin(2x) \cos(2x) + \cos(2x) \sin(2x)$.

Tip:

Memorizing product-to-sum formulas can help simplify trigonometric expressions, especially in integration problems!