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, \[ \boxed{ 1+\cos^22x = 2(\cos^4x+\sin^4x) } \]