Question

Verify the identity:
cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x

My work:
cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x
cos 4x + cos 2x = 2(1 - sin^2 2x - sin^2 x)
cos 4x + cos 2x = 2((2-cos 4x)/2 - (1-cos 2x)/2)
cos 4x + cos 2x = (2-cos 4x) - (1-cos 2x)

If this work is correct, could someone give me a step or two ahead and the reason for said step? If it isn't correct, could someone tell me where I went wrong here?

Answer 1

There is a much easier way of approaching this question. Would you like to know? @flixoe

Answer 2

Certainly.

Answer 3

First notice the identities:\[\bf \cos(2x)=1-2\sin^2(x) \implies \cos(4x)=1-2\sin^2(2x)\]Now let's work on the right side to make it like the left side using these identities:\[\bf R.S=2-2\sin^2(2x)-2\sin^2(x) = 1 + 1-2\sin^2(2x)-2\sin^2(x)\]\[\bf \implies R.S=(1-2\sin^2(2x))+(1-2\sin^2(x))=\cos(4x)+\cos(2x)=L.S\]
See how easy that was? I didn't have to do anything. Just make the 2 as 1 + 1, then re-arrange the expression and equate the identities and voila!

@flixoe

Answer 4

Oh I didn't understand that cos(4x) = 1 - 2sin^2(2x).

So:
cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x
cos 4x + cos 2x = 1 + 1 − 2sin^2(2x) − 2sin^2(x)
cos 4x + cos 2x = (1 - 2sin^2(2x)) + (1 - 2 sin^2(x)
cos 4x + cos 2x = cos 4x + cos 2x

Answer 5

yw =P