Prove the following identity: $$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
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$\begingroup$$$\cos (4\theta)=2\cos^2 (2\theta)-1=2(1-2\sin^2 \theta)^2-1=8\sin^4 \theta-8\sin^2 \theta+1$$
$$4\cos 2\theta = 4(1-2\sin^2 \theta)=4-8\sin^2 \theta$$
$\endgroup$ 0 $\begingroup$$$\cos 4\theta-4\cos 2\theta= (8\sin^4 \theta-8\sin^2 \theta+1)-(4-8\sin^2 \theta)=8\sin^4 \theta-3$$
$$\cos(4\theta)=2\cos^2(2\theta)-1\to\\ \cos(4\theta) - 4\cos(2\theta)=2\cos^2(2\theta)- 4\cos(2\theta)-1\to\\ \cos(4\theta) - 4\cos(2\theta)=2(1-2\sin^2(\theta))^2-4(1-2\sin^2(\theta))-1=8\sin^4(\theta)-3$$
$\endgroup$ $\begingroup$Since $\cos(4\theta)=2\cos^2(2\theta)-1$ and $\cos(2\theta)=1-\sin^2(\theta)$, we have $$\begin{align} \cos(4\theta) - 4\cos(2\theta)&=2\cos^2(2\theta)- 4\cos(2\theta)-1 \\ &=2(1-2\sin^2(\theta))^2-4(1-2\sin^2(\theta))-1\\ &=2-8\sin^2(\theta)+8\sin^4(\theta)-4+8\sin^2(\theta)-1\\ &=8\sin^4(\theta)-3. \end{align}$$
$\endgroup$ $\begingroup$LHS: $$ \cos (4 \theta)-4\cos(2\theta)=2\cos^2(2\theta)-1-4\cos(2\theta) $$ RHS: $$ 8\sin^4(\theta)-3=2\left( 2\sin^2 \theta\right)^2-3=2(1-\cos(2\theta))^2-3=2+2\cos^2(2\theta)-4\cos(2\theta)-3 $$
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