Can we express $\cos(4\theta)$ using only cosine?

Question

We can express $\cos(3\theta)$ using only cosines, can we do that with $4\theta$?

Answer 1

$$\cos ({4\theta}) = 2{\cos^2 ({2\theta})} - 1 = 2(2 \cos^2 \theta -1)^2 - 1$$

Answer 2

Hint:

$$\cos4t+\cos2t=2\cos3t\cos t$$

Now can you express $\cos3t,\cos2t$ in terms of $\cos t$

Answer 3

For all $n\in\Bbb N$, $\cos nt$ is a polynomial of degree $n$ in $\cos t$. These polynomials are Chebyshev polynomials. As an example $$\cos4t=2\cos^22t-1=2(2\cos^2t-1)^2-1$$ etc.

Answer 4

Yes, of course

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

$$2[2\cos ^2 (\theta)-1 ]^2-1 $$

$$2[4\cos ^4(\theta)-4\cos^2(\theta)+1 ]-1 $$

$$8\cos ^4 (\theta)-8\cos^2(\theta)+1 $$

Answer 5

Let $c=\cos\theta$ and $s\sin\theta$ \begin{align*} \cos 4\theta+i\sin4\theta&=(c+is)^4\\ &=c^4+4ic^3s-6c^2s^2-4ics^3+s^4 \end{align*}

Comparing the real parts,

\begin{align*} \cos 4\theta&=c^4-6c^2s^2+s^4\\ &=c^4-6c^2(1-c^2)+(1-c^2)^2\\ &=8c^4-8c^2+1 \end{align*}