To complete the proof showing that \( \sin^4 x = \frac{3 - 4 \cos 2x + \cos 4x}{8} \), let's go through the steps you've mentioned briefly and determine which expression fits into the fourth step.
- Starting from \( \sin^4 x = (\sin^2 x)^2 \).
- Then, using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \), we rewrite it as \( \sin^4 x = \left(\frac{1 - \cos 2x}{2}\right)^2 \).
- Simplifying that gives us: \[ \sin^4 x = \frac{(1 - \cos 2x)^2}{4} = \frac{1 - 2\cos 2x + \cos^2 2x}{4}. \]
In the following step, we want to manipulate \( \cos^2 2x \). We can use the identity \( \cos^2 2x = \frac{1 + \cos 4x}{2} \).
- Applying that identity gives us: \[ \sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4}. \]
Now we combine the terms in the numerator:
- The first term is \( 1 \).
- The second term is \( -2\cos 2x \).
- The third term is \( \frac{1 + \cos 4x}{2} \), which can also be written as \( \frac{2 + \cos 4x}{2} \).
Thus the numerator becomes: \[ = 1 - 2\cos 2x + \frac{2 + \cos 4x}{2}, \] = \( \frac{2 - 4\cos 2x + 2 + \cos 4x}{2} \) = \( \frac{4 - 4\cos 2x + \cos 4x}{2} \) = \( \frac{3 - 4\cos 2x + \cos 4x}{8} \).
Thus the expression that will complete the fourth step in the proof is: \[ \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4}. \] The correct choice is: \[ \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4}. \]