\( 2 \sin ^{2} \theta \cos ^{2} \theta+\cos ^{4} \theta=1-\sin ^{4} \theta \) 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. \[ 2 \sin ^{2} \theta \cos ^{2} \theta+\cos ^{4} \theta=\square \]
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Bonus Knowledge
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!