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\( 2 \sin ^{2} \theta \cos ^{2} \theta+\cos ^{4} \theta=1-\sin ^{4} \theta \) To verify the identity, work each side separately until you obtain the same expression. Start with the left side. Choose the correct step and transform the expression according to the step chosen. \[ 2 \sin ^{2} \theta \cos ^{2} \theta+\cos ^{4} \theta=\square \]

Ask by Gray Bartlett. in the United States
Feb 26,2025

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Answer

Both sides simplify to \( \cos^{2} \theta (1 + \sin^{2} \theta) \), so the identity is verified.

Solution

To verify the identity \( 2 \sin ^{2} \theta \cos ^{2} \theta+\cos ^{4} \theta=1-\sin ^{4} \theta \), we will start with the left side and simplify it step by step. Starting with the left side: \[ 2 \sin^{2} \theta \cos^{2} \theta + \cos^{4} \theta \] ### Step 1: Factor out \(\cos^{2} \theta\) We can factor out \(\cos^{2} \theta\) from the first term: \[ = \cos^{2} \theta (2 \sin^{2} \theta + \cos^{2} \theta) \] ### Step 2: Use the Pythagorean identity We know that \(\sin^{2} \theta + \cos^{2} \theta = 1\). Therefore, we can express \(2 \sin^{2} \theta + \cos^{2} \theta\) as follows: \[ = \cos^{2} \theta (2 \sin^{2} \theta + (1 - \sin^{2} \theta)) \] This simplifies to: \[ = \cos^{2} \theta (2 \sin^{2} \theta + 1 - \sin^{2} \theta) \] ### Step 3: Combine like terms Now, combine the terms inside the parentheses: \[ = \cos^{2} \theta (1 + \sin^{2} \theta) \] ### Step 4: Substitute back Now we can express this as: \[ = \cos^{2} \theta + \sin^{2} \theta \cos^{2} \theta \] ### Step 5: Use the Pythagorean identity again Using the identity \(\sin^{2} \theta + \cos^{2} \theta = 1\): \[ = \cos^{2} \theta + \sin^{2} \theta \cos^{2} \theta \] This is the left side simplified. Now we can express it as: \[ = \cos^{2} \theta (1 + \sin^{2} \theta) \] Now we can proceed to simplify the right side \(1 - \sin^{4} \theta\) to see if it matches. ### Right Side: Starting with the right side: \[ 1 - \sin^{4} \theta \] ### Step 1: Factor the expression We can factor \(1 - \sin^{4} \theta\) as a difference of squares: \[ = (1 - \sin^{2} \theta)(1 + \sin^{2} \theta) \] ### Step 2: Use the Pythagorean identity Using \(1 - \sin^{2} \theta = \cos^{2} \theta\): \[ = \cos^{2} \theta (1 + \sin^{2} \theta) \] ### Conclusion Now we have both sides simplified to: \[ \cos^{2} \theta (1 + \sin^{2} \theta) \] Thus, we have verified that: \[ 2 \sin^{2} \theta \cos^{2} \theta + \cos^{4} \theta = 1 - \sin^{4} \theta \] The identity holds true.

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Let's begin with the left side of the equation: \[ 2 \sin^{2} \theta \cos^{2} \theta + \cos^{4} \theta \] We can factor out \(\cos^{2} \theta\) from the first term, leading to: \[ \cos^{2} \theta (2 \sin^{2} \theta + \cos^{2} \theta) \] Next, we utilize the Pythagorean identity, \( \sin^{2} \theta + \cos^{2} \theta = 1 \): Substituting for \( 2 \sin^{2} \theta \) gives us: \[ 2 \sin^{2} \theta + \cos^{2} \theta = 2 (1 - \cos^{2} \theta) + \cos^{2} \theta = 2 - 2 \cos^{2} \theta + \cos^{2} \theta = 2 - \cos^{2} \theta \] Inserting back to our expression, we have: \[ \cos^{2} \theta (2 - \cos^{2} \theta) = 2 \cos^{2} \theta - \cos^{4} \theta \] Thus, the left side simplifies to: \[ 2 \cos^{2} \theta - \cos^{4} \theta \] Now we move on to the right side: \[ 1 - \sin^{4} \theta \] This can also be evaluated using the identity \( \sin^{4} \theta = (\sin^{2} \theta)^{2} \). By substituting \( \sin^{2} \theta = 1 - \cos^{2} \theta \), we get: \[ \sin^{4} \theta = (1 - \cos^{2} \theta)^{2} = 1 - 2\cos^{2} \theta + \cos^{4} \theta \] So, substituting back into our expression: \[ 1 - \sin^{4} \theta = 1 - (1 - 2\cos^{2} \theta + \cos^{4} \theta) = 2\cos^{2} \theta - \cos^{4} \theta \] Both sides now equal: \[ 2 \cos^{2} \theta - \cos^{4} \theta \] Thus, we have verified that \[ 2 \sin^{2} \theta \cos^{2} \theta+\cos^{4} \theta = 1-\sin^{4} \theta \] And the identity holds!

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