Respuesta :
[tex]\cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1[/tex]
using the identity [tex]\cos2x=\cos^2x-\sin^2x=2\cos^2x-1[/tex]. Expanding yields
[tex]\cos4x=8\cos^4x-8\cos^2x+1[/tex]
using the identity [tex]\cos2x=\cos^2x-\sin^2x=2\cos^2x-1[/tex]. Expanding yields
[tex]\cos4x=8\cos^4x-8\cos^2x+1[/tex]
[tex]\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
\boxed{2cos^2(\theta)-1}
\end{cases}
\\ \quad \\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\[/tex]
[tex]\bf cos[4x]\iff cos[2(2x)]\implies 2cos^2[2x]-1\iff 2[cos(2x)]^2-1 \\\\\\ 2[\ 2cos^2(x)-1\ ]^2-1\impliedby \textit{now, expanding the binomial} \\\\\\ 2[\ 2^2cos^4(x)-4cos^2(x)+1^2\ ]-1 \\\\\\ 2[\ 4cos^4(x)-4cos^2(x)+1\ ]-1 \\\\\\\ [\ 8cos^4(x)-8cos^2(x)+2\ ]-1 \\\\\\ 8cos^4(x)-8cos^2(x)+2-1\implies 8cos^4(x)-8cos^2(x)+1[/tex]
[tex]\bf cos[4x]\iff cos[2(2x)]\implies 2cos^2[2x]-1\iff 2[cos(2x)]^2-1 \\\\\\ 2[\ 2cos^2(x)-1\ ]^2-1\impliedby \textit{now, expanding the binomial} \\\\\\ 2[\ 2^2cos^4(x)-4cos^2(x)+1^2\ ]-1 \\\\\\ 2[\ 4cos^4(x)-4cos^2(x)+1\ ]-1 \\\\\\\ [\ 8cos^4(x)-8cos^2(x)+2\ ]-1 \\\\\\ 8cos^4(x)-8cos^2(x)+2-1\implies 8cos^4(x)-8cos^2(x)+1[/tex]