**Q : Which is 2*sin(4x)*cos(2x) written as a sum containing only sines?**

**A. sin 6x – sin 2x**

**B. sin 5x – sin3x**

**C. sin 5x + sin 3x**

**D. sin 6x + sin 2x**

A: This is another product-to-sum identity, only it is an the identity that involves sines and cosines mixed. The correct identity is:

sin(a)cos(b) = [sin(a + b) + sin(a – b)]/2

Multiply the 2 across to get:

2*sin(a)cos(b) = sin(a + b) + sin(a – b)

See how this now looks like our problem?

2*sin(4x)*cos(2x)!!

So, our a = 4x and our b = 2x… Plug those in to get:

2*sin(4x)*cos(2x) = sin(4x + 2x) + sin(4x – 2x)

2*sin(4x)*cos(2x) = sin(6x) + sin(2x)

**This is answer D**

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