Prove that \[ \cos4x=1-8\cos^2x+8\cos^4x \]
Proof:
Using the double angle identity:
\[
\cos4x=1-2\sin^22x
\]
Also,
\[
\sin2x=2\sin x\cos x
\]
Squaring both sides:
\[
\sin^22x=4\sin^2x\cos^2x
\]
Substituting:
\[
\cos4x
=
1-2(4\sin^2x\cos^2x)
\]
\[
=
1-8\sin^2x\cos^2x
\]
Using
\[
\sin^2x=1-\cos^2x
\]
we get
\[
\cos4x
=
1-8(1-\cos^2x)\cos^2x
\]
\[
=
1-8\cos^2x+8\cos^4x
\]
Hence proved,
\[
\boxed{
\cos4x=1-8\cos^2x+8\cos^4x
}
\]