Question

Simplify:
cos 3x
cos (2x+x)
cos (2x+x) = cos 2x cos x - sin 2x sin x
(cos^2(x) - sin^2(x))cos x - 2sin^2(x) cos x
(cos^3(x) - sin^2(x) cos x) - 2sin^2(x) cos x

Where do I go from here to reach one of these:

A)cos x - 4 cos x sin^2(x)

B) -sin^3(x) + 2 sin x cos x

C) -sin^2(x) + 2 sin x cos x

D) 2 sin^2(x) cos x - 2 sin x cos x

Answer 1

there are other "double angle" formulas you might try for \(\cos(2x)\) that are less cumbersome

Answer 2

like either \(1-2\sin^2(x)\) or \(2\cos^2(x)-1\)

Answer 3

Such as:
2cos^2(x) - 1
1 - 2sin^2(x)

So I just plug them in and hope it was the correct direction?

Answer 4

Then again seeing that the answers have high powers of sin, then it'd be a good idea to use 1 - 2sin^2(x)?

Answer 5

yeah that is what i would do

Answer 6

cos 3x
cos (2x+x)
cos (2x+x) = cos 2x cos x - sin 2x sin x
(1-2sin^2(x))cos x - 2sin^2(x) cos x
cos x - 2sin^2(x)cos x - 2sin^2(x) cos x
cos x - 4 cos x sin^2 x

Like so?