Question

Verify the equation:
cos(4x) = 1 - 6sin^2(x) + 4sin^4(x)
The answer I came up with is
cos(4x) = 1 - 8sin^2(x) + 8sin^4(x)
What am I doing wrong?

Answer 1

expansion of terms

Answer 2

Use half angle formula.

Answer 3

\[
\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}
\]

Answer 4

Oh wait, you're using double angle? \[
\cos(2\theta) = \cos^2(\theta)-\sin^2(\theta)
\]

Answer 5

yes ... double angles

Answer 6

You're probably just messing up your algebra.

Answer 7

You're probably right :) ... but I've reworked it several times

Answer 8

If I type out the work, will you take a look at it and see where I messed up?

Answer 9

yes

Answer 10

cos(4x)
= cos(2x + 2x)
= cos(2x)cos(2x) - sin(2x)sin(2x)
= [(cos^2(x) - sin^2(x)] [(cos^2(x) - sin^2(x)] - sin(2x)sin(2x)
= [1 - 2sin^2(x)][1 - 2sin^2(x)] - sin(2x)sin(2x)
= [1 - 4sin^2(x) + 4sin^4(x) - [2sin(x)cos(x) * 2sin(x)cos(x)]
= [1 - 4sin^2(x) + 4sin^4(x) - 4sin^2(x)cos^2(x)
= [1 - 4sin^2(x) + 4sin^4(x) - [4sin^2(x) * (1 - sin^2(x)]
= [1 - 4sin^2(x) + 4sin^4(x) - [4sin^2(x) - 4sin^4(x)]
= 1 - 8sin^2(x) + 8sin^4(x)
Whew ... :)

Answer 11

Looking over it now...

Answer 12

= [(cos^2(x) - sin^2(x)] [(cos^2(x) - sin^2(x)] - sin(2x)sin(2x)
= [1 - 2sin^2(x)][1 - 2sin^2(x)] - sin(2x)sin(2x)

This step doesn't make sense to me.

Answer 13

Wait nevermind, I get it now.

Answer 14

Seems that your work is correct.

Answer 15

:) Thank you so much for checking it for me! I really appreciate it!