Problem 40 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

Use trigonometric identities and algebraic simplification to prove the equation.

Step 1: Recall Trigonometric Identities

We will use the angle addition and subtraction formulas: 1. For cosine of sum: \( \cos(a + b) = \cos a \cos b - \sin a \sin b \)2. For cosine of difference: \( \cos(a - b) = \cos a \cos b + \sin a \sin b \).

Step 2: Apply the Formulas to the Problem

We want to calculate \( \cos(x+y) \) and \( \cos(x-y) \). Applying the formulas:\[ \cos(x+y) = \cos x \cos y - \sin x \sin y \]\[ \cos(x-y) = \cos x \cos y + \sin x \sin y \].

Step 3: Multiply the Trigonometric Expansions

Multiply the results from Step 2:\[(\cos x \cos y - \sin x \sin y)(\cos x \cos y + \sin x \sin y)\]This can be expanded using the difference of squares formula: \((a-b)(a+b) = a^2-b^2\).

Step 4: Simplify the Multiplication Expression

Using the difference of squares on the expression in Step 3:\[ (\cos x \cos y)^2 - (\sin x \sin y)^2\]Simplify to get \(\cos^2 x \cos^2 y - \sin^2 x \sin^2 y\).

Step 5: Use Pythagorean Trigonometric Identity

Recall the Pythagorean identity \( \cos^2 y + \sin^2 y = 1 \). Apply this:\[ \cos^2 x \cos^2 y - \sin^2 x \sin^2 y = (\cos^2 x)(1 - \sin^2 y) - \sin^2 y(\cos^2 x)\]The expression simplifies to:\[ \cos^2 x - \sin^2 y\], which proves the identity.

Key Concepts

Angle Addition Formula

The Angle Addition Formula is a crucial trigonometric identity used to find the cosine or sine of the sum of two angles. It's particularly useful when dealing with angles that aren't typical on the unit circle. In general, the formula for cosine is:

• For the sum of two angles: \( \cos(a + b) = \cos a \cos b - \sin a \sin b \)
• For the difference of two angles: \( \cos(a - b) = \cos a \cos b + \sin a \sin b \)

These formulas help break down complex trigonometric expressions into simpler components.

In our exercise, we are looking at \( \cos(x+y) \) and \( \cos(x-y) \). By plugging in values for \( x \) and \( y \), we apply the formula to transform the left-hand side of the equation into expressions that we can further manipulate.

This initially gives us, \( \cos x \cos y - \sin x \sin y \) and \( \cos x \cos y + \sin x \sin y \). These will be key as we look to simplify and prove the given identity.

Difference of Squares

The Difference of Squares is a powerful algebraic formula used to simplify products of sums and differences. It's written as \((a - b)(a + b) = a^2 - b^2\).

When faced with a product like the one in our exercise: \((\cos x \cos y - \sin x \sin y)(\cos x \cos y + \sin x \sin y)\), we can notice that if we let \(a = \cos x \cos y\) and \(b = \sin x \sin y\), it matches the \((a-b)(a+b)\) pattern.

Using the Difference of Squares, this becomes:

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

This transformation is critical as it simplifies the expression, allowing us to focus on the essence of the identity we are proving. It reduces the complexity and paves the way for substituting known values and identities to close out the proof.

Pythagorean Identity

The Pythagorean Identity is a fundamental principle in trigonometry that expresses the intrinsic relationship between sine and cosine. Given by \( \cos^2 \theta + \sin^2 \theta = 1 \), this identity holds true for every angle \( \theta \). It's often used to swap terms or simplify expressions in proofs or calculations.

In our exercise, we have reached an expression \( \cos^2 x \cos^2 y - \sin^2 x \sin^2 y \). By applying the Pythagorean Identity, we replace \( \cos^2 y \) with \(1 - \sin^2 y \). Automatically, the term \( \cos^2 x \cos^2 y \) becomes \( \cos^2 x(1 - \sin^2 y) \).

After distributing \( \cos^2 x \) and combining like terms, the expression simplifies neatly into \( \cos^2 x - \sin^2 y \), perfectly matching our original identity. This strategic use of the Pythagorean identity helps in bridging different parts of the expression, bringing us to a successful conclusion of the proof.