Answer
See the explanation below.
Work Step by Step
$\cos 4t=8{{\cos }^{4}}t-8{{\cos }^{2}}t+1$
Recall the double angle formula,
$\cos 2t={{\cos }^{2}}t-1$
Consider the left side of the given expression and apply the above formula,
$\begin{align}
& \cos 4t=\cos 2\left( 2t \right) \\
& =2{{\cos }^{2}}\left( 2t \right)-1 \\
& =2{{\left( 2{{\cos }^{2}}t-1 \right)}^{2}}-1
\end{align}$
Recall the algebraic square formula,
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Apply the algebraic square formula.
$\begin{align}
& \cos 4t=2\left( 4{{\cos }^{2}}t+1-4{{\cos }^{2}}t \right)-1 \\
& =\left( 8{{\cos }^{2}}t+2-8{{\cos }^{2}}t \right)-1 \\
& =8{{\cos }^{4}}t-8{{\cos }^{2}}t+1
\end{align}$
Hence, it is proved that the given identity $\cos 4t=8{{\cos }^{4}}t-8{{\cos }^{2}}t+1$ holds true.
