Prove the given trigonometric equation such that:
\[\cos 4x=1-8{{\sin }^{2}}x {{\cos }^{2}}x\]

Hint: For the given questions we will use the trigonometric identity as follows:
\[\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\]. So we can use the above formula with cos4x to get the required expression.

Complete step-by-step solution -
We have been asked to prove \[\cos 4x=1-8{{\sin }^{2}}x {{\cos }^{2}}x\].
We know that \[\cos 2A=2{{\cos }^{2}}A-1\].
Now taking left hand side \[=\cos 4x\]
\[\Rightarrow \cos 4x=\cos 2(2x)\]
Since it is in the form of cos2A, here \[A=2x\]
\[\Rightarrow \cos 4x=2{{\cos }^{2}}(2x)-1\]
We know that \[\cos 2x=2{{\cos }^{2}}x-1\]. So by substituting the values of cos2x in the above expression, we get as follows:
\[\Rightarrow \cos 4x=2{{\left( 2{{\cos }^{2}}x-1 \right)}^{2}}-1\]
By using \[{{\left( a-b \right)}^{2}}={{a}^{2}}+2ab-{{b}^{2}}\] we get as follows:
\[\begin{align}
  & \Rightarrow \cos 4x=2\left[ {{\left( 2{{\cos }^{2}}x \right)}^{2}}-2\left( 2{{\cos }^{2}}x \right).1+1 \right]-1 \\
 & =2\left[ {{\left( 2{{\cos }^{2}}x \right)}^{2}}-2\left( 2{{\cos }^{2}}x \right).1+1 \right]-1 \\
 & =2\left[ 4{{\cos }^{2}}x\left( {{\cos }^{2}}x-1 \right)+1 \right]-1 \\
\end{align}\]
Since we know that \[{{\cos }^{2}}x-1=-{{\sin }^{2}}x\]
\[\begin{align}
  & \Rightarrow \cos 4x=2\left[ 4{{\cos }^{2}}x\left( -{{\sin }^{2}}x \right)+1 \right]-1 \\
 & =2\left[ -4{{\cos }^{2}}x {{\sin }^{2}}x+1 \right]-1 \\
 & =-8{{\cos }^{2}}x {{\sin }^{2}}x+2-1 \\
 & =-8{{\cos }^{2}}x {{\sin }^{2}}x+1 \\
\end{align}\]
\[\Rightarrow \cos 4x=1-8{{\cos }^{2}}x {{\sin }^{2}}x=\] right hand side
Hence the left hand side is equal to the right hand side.
Therefore the given expression is proved.

Note: Be careful of the sign while doing calculation. Also don’t substitute the value of \[\left( {{\cos }^{2}}x-1 \right)\] is \[{{\sin }^{2}}x\] by mistake since we know that \[\left( {{\cos }^{2}}x-1 \right)\] is equal to \[\left( -{{\sin }^{2}}x \right)\] so take care of it. We can also prove the given trigonometric expression by considering the right hand side of the equation.