To complete the proof that shows \(\sin^4 x = \frac{3 - 4 \cos 2x + \cos 4x}{8}\), we can break down the proof step by step.
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Start with the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\).
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Then square both sides: \[ \sin^4 x = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{(1 - \cos 2x)^2}{4} \]
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Expanding \((1 - \cos 2x)^2\) gives: \[ \sin^4 x = \frac{1 - 2\cos 2x + \cos^2 2x}{4} \]
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Now, we can express \(\cos^2 2x\) in terms of \(\cos 4x\) using the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\). This gives us: \[ \cos^2 2x = \frac{1 + \cos 4x}{2} \]
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Substitute \(\cos^2 2x\) into the expression from step 3: \[ \sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} \]
To complete the fourth step of the proof, we need to combine like terms. This results in the expression: \[ \sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} \]
Now we can rewrite this as: \[ \sin^4 x = \frac{2 - 4\cos 2x + 1 + \cos 4x}{8} \] or simplified, \[ \sin^4 x = \frac{3 - 4\cos 2x + \cos 4x}{8} \]
The expression that will complete the fourth step before arriving at the final result is: \[ \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} \]
Looking at the provided options, the correct choice appears to be:
start fraction 1 minus 2 cosine 2 x plus start fraction 1 plus cosine 4 x over 2 end fraction over 4 end fraction
1 answer