Verify that the following equation is an identity. \[ (\cos 2 x+\sin 2 x)^{2}=1+\sin 4 x \] To verify the identity, start with the more complicated side and transform it to look like the other side. Ch \( (\cos 2 x+\sin 2 x)^{2} \) \( =\square \)

To verify the equation \( (\cos 2 x + \sin 2 x)^{2} = 1 + \sin 4 x \), we start with the left side: \[ (\cos 2 x + \sin 2 x)^{2} \] Using the binomial expansion: \[ = \cos^2 2x + 2 \cos 2x \sin 2x + \sin^2 2x \] We know from the Pythagorean identity that \( \cos^2 2x + \sin^2 2x = 1 \): \[ = 1 + 2 \cos 2x \sin 2x \] Now, using the double angle identity for sine, \( \sin 4x = 2 \sin 2x \cos 2x \): \[ = 1 + \sin 4x \] Thus, we have shown that: \[ (\cos 2 x + \sin 2 x)^{2} = 1 + \sin 4 x \] The left side and the right side match, confirming that the equation is indeed an identity!