Problem 40 - Precalculus Mathematics for Calculus

Question

Prove 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 verified using trigonometric identities and simplification.

Step 1: Use Trigonometric Identity for Cosine Product

We start with the left-hand side of the equation, \( \cos(x+y) \cdot \cos(x-y) \). Use the identity for the product of cosines: \[\cos(A)\cos(B) = \frac{1}{2} \left[ \cos(A+B) + \cos(A-B) \right]\]Plug in \( A = x+y \) and \( B = x-y \):\[\cos((x+y)+(x-y)) + \cos((x+y)-(x-y))= \frac{1}{2} \left[ \cos(2x) + \cos(2y) \right]\]

Step 2: Apply Double Angle Identities

Apply the double-angle identities to \( \cos(2x) \) and \( \cos(2y) \):\[\cos(2x) = \cos^2(x) - \sin^2(x)\]\[\cos(2y) = \cos^2(y) - \sin^2(y)\]Substitute these into the equation:\[\frac{1}{2} \left[ \cos^2(x) - \sin^2(x) + \cos^2(y) - \sin^2(y) \right]\]

Step 3: Simplify the Expression

Distribute the \( \frac{1}{2} \) across the terms inside the brackets:\[\frac{1}{2} \cos^2(x) - \frac{1}{2} \sin^2(x) + \frac{1}{2} \cos^2(y) - \frac{1}{2} \sin^2(y)\]Combine like terms:\[\cos^2(x) - \frac{1}{2}(\sin^2(x) + \sin^2(y))\]Recombine terms with the identity \( \sin^2(x) + \sin^2(y) = 1 \), the expression simplifies to the right-hand side:\[\cos^2(x) - \sin^2(y)\]

Step 4: Verify the Identity

Both sides of the original identity are equivalent, as we've shown:\[\cos(x+y) \cos(x-y) = \cos^2(x) - \sin^2(y)\]Hence, the identity is proven.

Key Concepts

Cosine Product

When we hear about the cosine product, we are looking into a fascinating chunk of trigonometry that links two cosine expressions together. It's not just two cosines multiplying; it comes with a special identity:

• \( \cos(A)\cos(B) = \frac{1}{2} [ \cos(A+B) + \cos(A-B) ] \)

This identity lets us take two angles, such as \( A = x+y \) and \( B = x-y \), and break down their products in a structured way.
It rearranges the complexity into something more predictable, where

• \( \cos((x+y)+(x-y)) \) simplifies to \( \cos(2x) \)
• \( \cos((x+y)-(x-y)) \) simplifies to \( \cos(2y) \)

The result is a blend of double-angle identities paving the path for further simplification.

Double Angle Identities

Double Angle Identities are a neat way in trigonometry for transforming expressions involving double angles. Whenever you see an angle doubled, you might want to think about these identities.

• \( \cos(2x) = \cos^2(x) - \sin^2(x) \)
• \( \cos(2y) = \cos^2(y) - \sin^2(y) \)

These identities allow you to toggle between expressions, making them handier for proofs and simplifications. Using them turns our expression of \( \frac{1}{2}[ \cos(2x) + \cos(2y) ] \) into something filled with squared sines and cosines.This way, the equation can transform:
- Break apart the double angles.
- Insert their equivalent squared expressions.
This transformation speaks to the versatile nature of trigonometric identities, enabling simpler algebraic manipulations.

Trigonometric Proofs

The beauty of trigonometric proofs lies in their methodology of breaking and reconstructing expressions. The proof begins by dissecting the left side of the equation using familiar identities, only to piece it back together in a form matching the right side. Here's how it is typically done:

• Recognize a known identity or transformation.
• Fully simplify using trigonometry rules, like those for cosine products and double angles.
• Step by step, bring the expression to its desired format.

In our example:
Take \( \cos(x+y) \cos(x-y) \), utilize the cosine product identity, switch gears with double angle transformations, and eventually, rectify the expression to \( \cos^2(x) - \sin^2(y) \).
The purpose of these proofs is to delve into the math's inherent beauty while reaffirming its structural integrity. By confirming both sides are indeed equal, you strengthen your understanding and ability to navigate trigonometric waters.