Problem 67 - Algebra and Trigonometry Real Mathematics, Real People

Question

Verify the identity. $$\cos (x+y) \cos (x-y)=\cos ^{2} x-\sin ^{2} y$$

Step-by-Step Solution

Answer

The identity \(\cos (x+y) \cos (x-y)=\cos ^{2} x-\sin ^{2} y\) is indeed a valid identity, which has been verified using the process explained.

Step 1: Recognize the Addition/Subtraction Formulas

We first recognize that the left-hand side of the given equation is in the form \(\cos (x+y) \cos (x-y)\), which suggests that the addition and subtraction formulas for cosine should be used. The addition formula is \(\cos (x \pm y)=\cos x \cos y \mp \sin x \sin y\), and we will be using this formula.

Step 2: Apply the Addition and Subtraction Formulas

We apply the addition and subtraction formulas to \(\cos (x+y) \cos (x-y)\), which results in \((\cos x \cos y - \sin x \sin y)(\cos x \cos y + \sin x \sin y)\).

Step 3: Simplify using Difference of Squares

The result from Step 2 is in a difference of squares form, \(a^2 - b^2\), which can be simplified using the formula \(a^2 - b^2 = (a-b)(a+b)\). This gives us \(\cos^2x - \sin^2y\).

Step 4: Verification

Since \(\cos^2x - \sin^2y\) is exactly the same as the right-hand side of the original equation, the identity is verified.