Problem 45 - Precalculus: Functions and Graphs

Question

Exer. \(37-46:\) Verify the identity. $$\frac{1}{\cot \alpha-\cot \beta}=\frac{\sin \alpha \sin \beta}{\sin (\beta-\alpha)}$$

Step-by-Step Solution

Answer

The identity is verified.

Step 1: Simplify the Left Side

The expression on the left side is \( \frac{1}{\cot \alpha - \cot \beta} \). Start by expressing cotangents in terms of sine and cosine: \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \) and \( \cot \beta = \frac{\cos \beta}{\sin \beta} \). Thus, \( \cot \alpha - \cot \beta = \frac{\cos \alpha}{\sin \alpha} - \frac{\cos \beta}{\sin \beta} \).

Step 2: Combine the Fractions

Combine \( \frac{\cos \alpha}{\sin \alpha} - \frac{\cos \beta}{\sin \beta} \) into a single fraction: \[ \frac{\cos \alpha \sin \beta - \cos \beta \sin \alpha}{\sin \alpha \sin \beta} \].

Step 3: Simplify Further Using Trigonometric Identities

The numerator \( \cos \alpha \sin \beta - \cos \beta \sin \alpha \) can be simplified using the sine difference identity: \( \cos \alpha \sin \beta - \cos \beta \sin \alpha = \sin(\beta - \alpha) \).

Step 4: Take the Reciprocal

Substitute \( \sin(\beta - \alpha) \) for the numerator in the fraction: \[ \frac{\sin(\beta - \alpha)}{\sin \alpha \sin \beta} \]. Taking the reciprocal, we gain \( \frac{\sin \alpha \sin \beta}{\sin(\beta - \alpha)} \).

Step 5: Compare with the Right Side

The expression is now \( \frac{\sin \alpha \sin \beta}{\sin(\beta - \alpha)} \), which matches the right side of the given identity, confirming the identity is verified.

Key Concepts

Sine Difference Identity

The Sine Difference Identity is a fundamental trigonometric identity that simplifies expressions involving the sine function of the difference of two angles. The formula is given by:

• \[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]

This identity helps in verifying trigonometric equations and transformations. In the provided exercise, this identity was instrumental in simplifying the numerator \( \cos \alpha \sin \beta - \cos \beta \sin \alpha \). By recognizing this expression as \( \sin(\beta - \alpha) \), we were able to transform the original equation into a simpler form, leading directly to the verification of the identity. Understanding how to use the sine difference identity is crucial for tackling a wide range of trigonometric problems. It showcases how identities can make simplification more manageable and elegant.

Cotangent

The cotangent function is one of the lesser-used trigonometric functions, compared to sine, cosine, and tangent. It is defined as the reciprocal of the tangent function:

• \[ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \]

This gives us a unique way to express relationships in trigonometry. In the exercise, we used the definition of cotangent to re-express the difference \( \cot \alpha - \cot \beta \) in terms of sine and cosine. This step was key in reaching a common denominator, which then allowed us to use the sine difference identity. Having a strong grasp of the cotangent’s role as a reciprocal of tangent helps in understanding its relationships with sine and cosine.

Trigonometric Simplification

Trigonometric simplification involves reducing complex trigonometric expressions to simpler forms. This is done using identities and algebraic manipulation. In the solution to the provided exercise, the main simplification process involved rewriting the terms in a way that allowed the identities to be applied efficiently.

• Transforming \( \cot \alpha - \cot \beta \) into a single fraction was a crucial step.
• Recognizing the numerator as a sine difference identity further simplified the work.

This process highlights the importance of recognizing patterns and applying identities not just mechanically, but strategically. Simplifying expressions not only confirms identities but also boosts understanding of the connections between trigonometric functions.

Reciprocal Identities

Reciprocal identities are basic trigonometric identities that express the main trigonometric functions as reciprocals of one another. For example:

• \( \tan \theta = \frac{1}{\cot \theta} \)
• \( \cot \theta = \frac{1}{\tan \theta} \)
• \( \sec \theta = \frac{1}{\cos \theta} \)
• \( \csc \theta = \frac{1}{\sin \theta} \)

These identities are vital in transforming and simplifying expressions involving trigonometric functions. In the exercise, understanding that \( \cot \alpha \) can be expressed as \( \frac{1}{\tan \alpha} \) helped in re-expressing the problem in terms of sine and cosine. This step perpetuates simplification by breaking down complex ratios into more manageable components using basic operations. Reciprocity is a potent tool in the algebra of trigonometry.