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 \( (\cos 2 x+\sin 2 x)^{2} \) \( =\square \)

Sure, let's verify the trigonometric identity: \[ (\cos 2x + \sin 2x)^{2} = 1 + \sin 4x \] **Step 1: Expand the Left-Hand Side (LHS)** Start with the more complicated side, which is the left-hand side (LHS): \[ (\cos 2x + \sin 2x)^{2} \] Use the algebraic identity \((a + b)^2 = a^2 + 2ab + b^2\) to expand the expression: \[ (\cos 2x + \sin 2x)^{2} = \cos^2 2x + 2 \cos 2x \sin 2x + \sin^2 2x \] **Step 2: Simplify Using Trigonometric Identities** 1. **Combine \(\cos^2 2x\) and \(\sin^2 2x\):** Recall the Pythagorean identity: \[ \cos^2 \theta + \sin^2 \theta = 1 \] Applying this to \(\theta = 2x\): \[ \cos^2 2x + \sin^2 2x = 1 \] 2. **Simplify the Middle Term:** The middle term is \(2 \cos 2x \sin 2x\). Use the double-angle identity for sine: \[ 2 \sin \theta \cos \theta = \sin 2\theta \] Applying this to \(\theta = 2x\): \[ 2 \cos 2x \sin 2x = \sin 4x \] **Step 3: Combine the Simplified Terms** Now, substitute the simplified terms back into the expanded expression: \[ \cos^2 2x + 2 \cos 2x \sin 2x + \sin^2 2x = 1 + \sin 4x \] **Conclusion** Thus, we have: \[ (\cos 2x + \sin 2x)^{2} = 1 + \sin 4x \] This confirms that the given equation is indeed an identity.