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\(cos(4x)=1-8sin^2(x)cos^2(x)\)

I know that I need to start the problem on the left, like this:

\(cos(2(2x))\)

Then treat 2x like it equals θ.

\(cos(2\theta)\)

Then I use a double angle formula, but since there are 3 for cosine, I don't know which one to use.

 Apr 6, 2018
 #1
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Misread the question! Sorry.

 Apr 6, 2018
edited by Guest  Apr 6, 2018
 #2
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Verify the following identity:
cos(4 x) = 1 - 8 sin(x)^2 cos(x)^2

cos(4 x) = cos(4 x) = 2 cos(2 x)^2 - 1:


2 cos(2 x)^2 - 1 = ^?1 - 8 cos(x)^2 sin(x)^2

cos(2 x) = 2 cos(x)^2 - 1:


2 (2 cos(x)^2 - 1)^2 - 1 = ^?1 - 8 cos(x)^2 sin(x)^2

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


2 1 - 4 cos(x)^2 + 4 cos(x)^4 - 1 = ^?1 - 8 cos(x)^2 sin(x)^2

2 (1 - 4 cos(x)^2 + 4 cos(x)^4) - 1 = 1 - 8 cos(x)^2 + 8 cos(x)^4:


1 - 8 cos(x)^2 + 8 cos(x)^4 = ^?1 - 8 cos(x)^2 sin(x)^2

sin(x)^2 = 1 - cos(x)^2:


1 - 8 cos(x)^2 + 8 cos(x)^4 = ^?1 - 8 cos(x)^2 1 - cos(x)^2

-8 cos(x)^2 (1 - cos(x)^2) = 8 cos(x)^4 - 8 cos(x)^2:
1 - 8 cos(x)^2 + 8 cos(x)^4 = ^?8 cos(x)^4 - 8 cos(x)^2 + 1

 

The left-hand side and right-hand side are identical:                                                                   (identity has been verified)

 Apr 6, 2018

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