+895 \(cos(4x)=1-8sin^2(x)cos^2(x)\)
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.
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)
