Problem 46 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The proof is complete. The given expression \(\sin(\alpha+\beta)\sin(\alpha-\beta)\) is indeed equal to \(\cos^2\beta - \cos^2\alpha\).

Step 1: Apply Sum-to-Product Formulas

The problem involves trigonometric identities, specifically the sum and difference in sine. To simplify, let's first use sum-to-product formulas like this: Apply the formulas \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\) and \(\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\).

Step 2: Expand and simplify

The next step is to expand and simplify. Multiply \(\sin(\alpha + \beta)\) by \(\sin(\alpha - \beta)\), which results in: \((\sin \alpha \cos \beta + \cos \alpha \sin \beta) * (\sin \alpha \cos \beta - \cos \alpha \sin \beta)\). Utilize the formula \((a+b)*(a-b) = a^2 - b^2\), where \(a = \sin \alpha \cos \beta\) and \(b = \cos \alpha \sin \beta\). This results in: \(\sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta\).

Step 3: Use Pythagorean Identity

The final step is to apply the Pythagorean identity for cosines. The Pythagorean identities are forms of the Pythagorean Theorem, which allows us to replace \(\cos^2 \alpha\) and \(\cos^2 \beta\) using the identity \(\cos^2 x = 1 - \sin^2 x\). Thus the previous expression becomes: \(\sin^2 \alpha (1 - \sin^2 \beta) - (1 - \sin^2 \alpha) \sin^2 \beta\) which simplifies to \(\cos^2 \beta - \cos ^2 \alpha\), proving the given trigonometric identity.