The expression \( 100 \sin(2t) \cos(4t) \) can be simplified using the product-to-sum identities. The relevant identity for the product of sine and cosine is:

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

In your case, \( A = 2t \) and \( B = 4t \). Plugging these values into the identity gives:

\[
\sin(2t)\cos(4t) = \frac{1}{2}[\sin(2t + 4t) + \sin(2t - 4t)]
\]

Simplifying the arguments inside the sine functions:

\[
\sin(2t)\cos(4t) = \frac{1}{2}[\sin(6t) + \sin(-2t)]
\]

Since \( \sin(-x) = -\sin(x) \):

\[
\sin(2t)\cos(4t) = \frac{1}{2}[\sin(6t) - \sin(2t)]
\]

Now, multiplying by 100 to account for the original coefficient:

\[
100\sin(2t)\cos(4t) = 50[\sin(6t) - \sin(2t)]
\]

So the simplified expression is:

\[
\boxed{50\sin(6t) - 50\sin(2t)}
\]

Would you like more details or have any questions about this process?

Here are some related questions you might consider:

1. How can the product-to-sum identities be used in solving integrals of trigonometric functions?
2. What is the product-to-sum identity for the product of two sine functions?
3. How can the expression \( \cos(3t)\cos(5t) \) be simplified using a similar identity?
4. What role do these identities play in simplifying Fourier transforms?
5. Can you solve for the integral of \( \sin(2t)\cos(4t) \) over a specific interval?

**Tip:** Mastering product-to-sum identities is essential for simplifying complex trigonometric expressions and is particularly helpful in solving problems in calculus and physics.

Learn how to simplify the trigonometric expression 100sin(2t)cos(4t) step-by-step using product-to-sum identities. This problem involves applying mathematical identities to simplify complex trigonometric expressions, suitable for advanced high school and college students.

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