Prove that $4\cos^4x-2\cos2x-1/2\cos4x$ is independent of x $4\cos^4x-2\cos2x-1/2\cos4x$
I don't know how to proceed.  $\cos4x=2\cos^2(2x)-1$ Is this useful at all?
 A: Yes, $\cos4x=2\cos^2(2x)-1$ is useful, also $\cos^2x=\dfrac{1+\cos2x}{2}$.
Now
$$4\cos^4x-2\cos2x-1/2\cos4x=4(\dfrac{1+\cos2x}{2})^2-2\cos2x-1/2(2\cos^2(2x)-1)$$
then let $\cos2x=k$. Can you proceed?
A: $$4\cos^4x-2\cos2x-\frac{1}{2}\cos4x=(1+\cos2x)^2-2\cos2x-\frac{1}{2}(2\cos^22x-1)=\frac{3}{2}.$$
A: Hint: using $\cos2x = 1-2\sin^2(x)$, $\sin(2x) = 2\sin(x)\cos(x)$ and $\sin^2(x) + \cos^2(x) = 1$ to get the following and simplify it:
$$S = 4\cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4\cos^4(x)-2+4\sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4\cos^4(x)-2+1/2 + 4\sin^2(x)(1-\cos^2(x)) = 4\cos^4(x)-2+1/2 + 4\sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
A: Yes but an alternative is to use the definition in terms of exponentials
$$
\cos x=\frac{1}{2}(e^{ix}+e^{-ix})
$$
So
$$
\cos^4 x=\frac{1}{2^4}(e^{ix}+e^{-ix})^4
$$
If you expand the right-hand side with the binomial theorem and re-apply the exponential definition of $\cos$ e.g. $\cos {2x} = \frac{1}{2}(e^{i2x}+e^{-i2x})$ the result appears.
