To rewrite $\cos(4x)$ in terms of $\cos(2x)$, we use the double-angle formula for cosine:

$\cos(2\theta) = 2\cos^2(\theta) - 1$

Here, let $\theta = 2x$, so $\cos(4x) = \cos(2 \cdot 2x)$. Applying the double-angle formula:

$\cos(4x) = 2\cos^2(2x) - 1$

Thus, $\cos(4x)$ is expressed in terms of $\cos(2x)$ as:

$\cos(4x) = 2\cos^2(2x) - 1$

Let me know if you'd like further clarification or examples!

Related Questions:

1. How can you derive $\cos(6x)$ using similar techniques?
2. How does $\cos^2(x)$ relate to $\cos(2x)$?
3. What is the equivalent expression for $\sin(4x)$ in terms of $\sin(2x)$?
4. How can the double-angle formula for cosine simplify trigonometric integrals?
5. How do trigonometric identities like this apply to Fourier analysis?

Tip:

When rewriting expressions using trigonometric identities, always check for simplifications by using other related formulas, such as $\sin^2(x) + \cos^2(x) = 1$.