To write the product $\sin(x) \sin(3x)$ as a sum, we can use the trigonometric identity for the product of sines:

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

For this problem:

• $A = x$
• $B = 3x$

Now, substitute these values into the identity:

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

Simplify inside the cosine functions:

1. $x - 3x = -2x$
2. $x + 3x = 4x$

So, we have:

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

Since $\cos(-2x) = \cos(2x)$ (because cosine is an even function), this further simplifies to:

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

Thus, the product $\sin(x) \sin(3x)$ written as a sum is:

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

Would you like additional details or have any questions about this?

Here are some related questions to expand your understanding:

1. How would you write $\sin(x) \sin(5x)$ as a sum?
2. What is the identity for the product of two cosine functions?
3. Can you derive the product-to-sum identities from the angle addition formulas?
4. How does the even-odd property of trigonometric functions affect simplifications?
5. What is the product-to-sum identity for $\cos(A) \sin(B)$?

Tip: When using product-to-sum identities, always simplify by checking even-odd properties of the trigonometric functions involved.