Problem 45 - Precalculus Essentials

Question

Verify each identity. $$\cos (\alpha+\beta) \cos (\alpha-\beta)=\cos ^{2} \beta-\sin ^{2} \alpha$$

Step-by-Step Solution

Answer

The identity has been successfully verified, the left side of the equation simplifies to exactly match the right side, \( \cos^{2}(\beta) - \sin^{2}(\alpha) \).

Step 1: Express LHS in terms of sin and cos

The first step is to apply the cosine difference and cosine addition identity formulas to the left-hand side (LHS): \( \cos(\alpha + \beta) \) and \( \cos(\alpha - \beta) \). These formulas are: \( \cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b) \). Let's apply these formulas to our equation. So, \( \cos(\alpha + \beta) \cos(\alpha - \beta) = [\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)][\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)]

Step 2: Simplify LHS

The second step is to simplify the expression obtained from step 1 by applying the multiplication of binomials formula \( (a-b)(a+b) = a^{2} - b^{2} \). This yields \( \cos^{2}(\beta) - \sin^{2}(\alpha) \) which is the RHS of the equation.

Step 3: Verification

Lastly, we compare the simplified LHS with the RHS of our given equation. Both sides are \( \cos^{2}(\beta) - \sin^{2}(\alpha) \), thus the original equation is an identity and has been verified.