Cosine Simplification

Yahoo Answers is definitely the place everyone goes to find an answer for a math question. Otherwise, there is no explanation for the kinds of questions you can find there. Today's question is about simplification of a trigonometric expression:

Simplify: (cos 3x)(1 - cos 2x + cos 4x - cos 6x)

To solve that, we will need some major identities: the sum of cosines, the difference of cosines, the sum of sines, and the product of sines and cosines. With these formulas, this question can be easily solved.

Let's start with the formulas:
cos α + cos β = 2cos((α + β)/2)cos((α - β)/2)
cos α - cos β = -2sin((α + β)/2)sin((α - β)/2)
sin α + sin β = 2sin((α + β)/2)cos((α - β)/2)
sinαcosβ = 1/2(sin(α + β)sin(α - β))

Now, to avoid complications, let's aim for the average. We can turn 1 into cos 0, so we have the cosines of 0 and 6x and the cosines of 2x and 4x. In both formulas they are divided by two, so they should probably cancel each other.

cos 0 - cos 6x = -2sin((6x + 0) / 2)sin(6x - 0)/2 = -2sin² 3x
cos 4x - cos 2x = -2sin((4x + 2x)/2)sin((4x - 2x)/2) = -2sin 3x * sinx

In conclusion, the problem has just become:
(cos 3x)(-2sin² 3x - 2sin 3x * sin x)

Distribute:
-2sin 3x * sin 3x * cos 3x - 2sin 3x * cos 3x * sinx

Transform into a sum of a cosine and a sine (I picked α and β to be 3x, because there will be a sine of zero, which is zero):
-2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin 3x - 2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin x
-sin 6x * sin 3x - sin 6x * sin x

We are getting really close now, just hold on:
-(sin 6x)(sin 3x + sin x)
-(sin 6x)(2sin((3x + x)/2)cos((3x - x)/2)
-2sin 6x * sin 2x * cos x

And that's it.

Nadav

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