Prove that \[ 1+\cos^22x=2(\cos^4x+\sin^4x) \]
Proof:
Start with the right-hand side:
\[
RHS=2(\cos^4x+\sin^4x)
\]
Using the identity
\[
a^2+b^2=(a+b)^2-2ab
\]
we get
\[
\cos^4x+\sin^4x
=
(\cos^2x+\sin^2x)^2
-2\sin^2x\cos^2x
\]
Using
\[
\sin^2x+\cos^2x=1
\]
therefore,
\[
\cos^4x+\sin^4x
=
1-2\sin^2x\cos^2x
\]
Hence,
\[
RHS
=
2\left(1-2\sin^2x\cos^2x\right)
\]
\[
=
2-4\sin^2x\cos^2x
\]
Using
\[
\sin2x=2\sin x\cos x
\]
Squaring both sides:
\[
\sin^22x=4\sin^2x\cos^2x
\]
Therefore,
\[
RHS
=
2-\sin^22x
\]
Using
\[
\sin^2\theta+\cos^2\theta=1
\]
we get
\[
\sin^22x=1-\cos^22x
\]
Substituting:
\[
RHS
=
2-(1-\cos^22x)
\]
\[
=
1+\cos^22x
\]
\[
= LHS
\]
Hence proved,
\[
\boxed{
1+\cos^22x
=
2(\cos^4x+\sin^4x)
}
\]