**Q: Which of the following expresses 2cos(5x)* cos(2x) as a sum containing only sines or cosines?**

**A. cos(7x)-cos(3x)**

**B. cos(6x)+cos(4x)**

**C. cos(7x)+cos(3x)**

**D. cos(6x)-cos(4x)**

A: Use the **product-to-sum identity**, which states:

cos(a)*cos(b) = [cos(a-b) + cos(a + b)] / 2

Multiply the 2 over to get:

2cos(a)*cos(b) = cos(a-b) + cos(a + b)

Notice that this looks **a lot** like our problem?

2cos(5x)* cos(2x)…..

Therefore, our a = 5x and our b = 2x:

2cos(5x)*cos(2x) = cos(5x – 2x) + cos(5x + 2x)

2cos(5x)*cos(2x) = cos(3x) + cos(7x)

**This is answer C** (just written in flipped order, which is fine!)

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