Problem 43 - Algebra and Trigonometry with Analytic Geometry

Question

Use fundamental identities to write the first expression in terms of the second, for any acute angle \(\theta\). $$\cot \theta, \sin \theta$$

Step-by-Step Solution

Answer

\(\cot \theta = \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta}\)

Step 1: Understanding the Relationship

To express \(\cot \theta\) in terms of \(\sin \theta\), we need to use fundamental trigonometric identities. We begin by recalling the basic identity that relates cotangent and sine, which is \(\cot \theta = \frac{\cos \theta}{\sin \theta}\). So, the next task is to express \(\cos \theta\) in terms of \(\sin \theta\).

Step 2: Using the Pythagorean Identity

We use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Solving for \(\cos \theta\), we have \(\cos \theta = \sqrt{1 - \sin^2 \theta}\). This expression allows us to directly substitute in terms of \(\sin \theta\).

Step 3: Substitute into Cotangent Formula

Substitute \(\cos \theta = \sqrt{1 - \sin^2 \theta}\) into the cotangent formula \(\cot \theta = \frac{\cos \theta}{\sin \theta}\). This gives us: \(\cot \theta = \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta}\).

Step 4: Simplify the Expression

There are no further simplifications possible that can reduce the expression \(\frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta}\) into a simpler form using the given elementary functions, as it is already in its simplest form in terms of \(\sin \theta\).

Key Concepts

Cotangent and Sine Relationship

Understanding the relationship between cotangent and sine is fundamental in trigonometry. Cotangent is the reciprocal of the tangent function, and it connects closely with sine and cosine. When considering \(\cot \theta\), it's important to remember that it is defined as:

• \(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

This means \(\cot \theta\) depends on both sine and cosine. Our goal is to express it solely in terms of \(\sin \theta\). Knowing this relation helps us bridge \(\cot \theta\) to other trigonometric functions, especially when working on exercises where direct computation of cotangent may not be possible. Understanding this identity allows one to solve various trigonometric problems by leveraging the relationships between the basic trigonometric functions.

Pythagorean Identity

The Pythagorean identity is a cornerstone in trigonometry, reflecting the relationship between sine and cosine in a simple equation:

• \(\sin^2 \theta + \cos^2 \theta = 1\)

This fundamental identity allows us to derive one trigonometric function from another. If we know the value of the sine of an angle, we can determine the cosine by rearranging the identity:

• \(\cos \theta = \sqrt{1 - \sin^2 \theta}\)

Understanding this identity is crucial for solving problems where expressing one function in terms of another is necessary. It provides the tools needed to navigate between sine and cosine seamlessly, especially when expressing \(\cot \theta\) in terms of \(\sin \theta\). This identity turns the abstract relationships between the functions into more tangible expressions that are easier to manipulate and understand in the context of a problem.

Expressing Cosine in Terms of Sine

Expressing cosine solely in terms of sine involves leveraging the Pythagorean identity. Once we have the Pythagorean relationship, deriving cosine becomes straightforward:

• Given the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Rearrange to solve for cosine: \(\cos^2 \theta = 1 - \sin^2 \theta\)

Thus, the expression for cosine in terms of sine is:

• \(\cos \theta = \sqrt{1 - \sin^2 \theta}\)

Utilizing this relationship is key in problems that involve substituting \(\cos \theta\) in trigonometric identities. It simplifies computations when you need to express everything in terms of \(\sin \theta\), such as when calculating \(\cot \theta\) in terms of sine. This transition from cosine to sine is often employed to maintain consistency in function terms and ensure expressions remain within a single trigonometric function, making complex problems easier to solve.