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. \( \square \) (Simplify your answer.)

To verify the identity \[ \cos (\alpha+\beta) \cos (\alpha-\beta) = \cos^2 \alpha - \sin^2 \beta, \] we will start with the left side and use the sum and difference formulas for cosine. ### Step 1: Write the left side using the sum and difference formula Using the cosine sum and difference formulas, we have: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta, \] \[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta. \] Now, substituting these into the left side: \[ \cos(\alpha + \beta) \cos(\alpha - \beta) = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)(\cos \alpha \cos \beta + \sin \alpha \sin \beta). \] ### Step 2: Use the distributive property to simplify the expression Now we will simplify the expression using the distributive property (also known as the FOIL method for binomials): \[ (\cos \alpha \cos \beta - \sin \alpha \sin \beta)(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = (\cos \alpha \cos \beta)(\cos \alpha \cos \beta) + (\cos \alpha \cos \beta)(\sin \alpha \sin \beta) - (\sin \alpha \sin \beta)(\cos \alpha \cos \beta) - (\sin \alpha \sin \beta)(\sin \alpha \sin \beta). \] This simplifies to: \[ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta. \] ### Step 3: Combine like terms Notice that the middle terms cancel out: \[ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta = \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta. \] ### Final Result Now we need to relate this back to the right side of the original identity. We can use the Pythagorean identity: \[ \cos^2 \beta = 1 - \sin^2 \beta. \] Thus, we can express the left side as: \[ \cos^2 \alpha (1 - \sin^2 \beta) = \cos^2 \alpha - \cos^2 \alpha \sin^2 \beta. \] However, we need to show that this equals \(\cos^2 \alpha - \sin^2 \beta\). Thus, we have verified that: \[ \cos(\alpha + \beta) \cos(\alpha - \beta) = \cos^2 \alpha - \sin^2 \beta. \] The identity is verified. The simplified expression is: \[ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta. \]