Verify the identity. \( \cos (\alpha+\beta) \cos (\alpha-\beta)=\cos ^{2} \alpha-\sin ^{2} \beta \) Write the left side of the identity using the sum and difference formula. ( \( \cos \alpha \cos \beta-\sin \alpha \sin \beta)(\cos \alpha \cos \beta+\sin \alpha \sin \beta \) ) (Type the terms of your expression in the same order as they appear in the original expression. Use the distributive property to simplify the expression from the previous step. \( \cos ^{2} \alpha \cos ^{2} \beta-\sin ^{2} \alpha \sin ^{2} \beta \) (Simplify your answer.) Substitute \( 1-\sin ^{2} \beta \) for \( \cos ^{2} \beta \) in the expression from the previous step and distribute. \( \square \)

To verify the identity, let's start with the left side: Using the sum and difference formula, we have: \[ \cos (\alpha+\beta) \cos (\alpha-\beta) = \left( \cos \alpha \cos \beta - \sin \alpha \sin \beta \right) \left( \cos \alpha \cos \beta + \sin \alpha \sin \beta \right) \] Now, applying the distributive property, we expand this: \[ = \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta \] Next, we recognize that \(\cos^2 \beta\) can be rewritten using the Pythagorean identity: \[ \cos^2 \beta = 1 - \sin^2 \beta \] So substituting \(1 - \sin^2 \beta\) for \(\cos^2 \beta\): \[ = \cos^2 \alpha (1 - \sin^2 \beta) - \sin^2 \alpha \sin^2 \beta \] Distributing \(\cos^2 \alpha\): \[ = \cos^2 \alpha - \cos^2 \alpha \sin^2 \beta - \sin^2 \alpha \sin^2 \beta \] Now, we combine terms to verify they match the right side, which is \(\cos^2 \alpha - \sin^2 \beta\). Thus, the original identity is confirmed as true. The next step where we denote it is fully expanded: \[ = \cos^2 \alpha - \sin^2 \beta (\cos^2 \alpha + \sin^2 \alpha) \] And as per the identity \(\cos^2 \alpha + \sin^2 \alpha = 1\), it simplifies to: \[ = \cos^2 \alpha - \sin^2 \beta \] This shows the verification is complete. \( \square \)