Problem 51 - Algebra 2

Question

Verify each identity. Express \(\cos \theta \csc \theta \cot \theta\) in terms of \(\sin \theta\)

Step-by-Step Solution

Answer

The expression \(\cos \theta \csc \theta \cot \theta\) in terms of \(\sin \theta\) is \(\frac{1 - \sin^2 \theta}{\sin^2 \theta}\).

Step 1: Break down the given function

Start by breaking down the given trigonometric function \(\cos \theta \csc \theta \cot \theta\) into their definitions. The definitions are as follows: \(\cos \theta = \cos \theta\), \(\csc \theta = \frac{1}{\sin \theta}\), and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).

Step 2: Substitute the definitions

Substitute the defined functions back into the given function. So, \(\cos \theta \csc \theta \cot \theta\) becomes \(\cos \theta * \frac{1}{\sin \theta} * \frac{\cos \theta}{\sin \theta}\). Simplifying this, you get \(\frac{\cos^2 \theta}{\sin^2 \theta}\).

Step 3: Apply Pythagorean Identity

Finally, replace \(\cos^2 \theta\) with \(1 - \sin^2 \theta\) by using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). This gives you \(\frac{1 - \sin^2 \theta}{\sin^2 \theta}\).

Key Concepts

Pythagorean identity

When working with trigonometric identities, a fundamental concept is the Pythagorean identity. It provides a relationship between the primary trigonometric functions: sine and cosine. The Pythagorean identity is expressed as:\[ \sin^2 \theta + \cos^2 \theta = 1 \]This identity holds for all values of \( \theta \) and is an essential tool in simplifying trigonometric expressions. In the context of our problem, it allows us to substitute \( \cos^2 \theta \) with \( 1 - \sin^2 \theta \).

• It helps convert expressions, making them simpler to relate and work with.
• Offers a pathway to express one function in terms of another, enhancing problem-solving strategies.

Remember, by leveraging the Pythagorean identity, you can facilitate the transition from a complex expression to a more standard form, often involving a single trigonometric function.

trigonometric functions

Trigonometric functions are the foundation of solving trigonometric identities and equations. The three primary trigonometric functions involved in this exercise are sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and their reciprocals, cosecant (\( \csc \theta = \frac{1}{\sin \theta} \)), and cotangent (\( \cot \theta = \frac{\cos \theta}{\sin \theta} \)). Here are some key points:

• Cosine: Represents the horizontal component of a point on the unit circle.
• Cosecant: It's the reciprocal of sine and relates to the relative position of the angle.
• Cotangent: Computed as the ratio of cosine to sine, providing an inter-function relationship.

Understanding these relationships helps in converting and simplifying trigonometric expressions, such as in the original task where these functions are manipulated into an expression of \( \sin \theta \). The ability to shift between these functions facilitates verification and solution of identities.

simplifying expressions

Simplifying trigonometric expressions is crucial for making challenging problems more manageable and understandable. The key is to break down complex terms into basic identities and relations. Let's consider the simplification task given:1. **Break down the expression:** Convert each component of \( \cos \theta \csc \theta \cot \theta \) into its fundamental definitions.2. **Substitute and simplify:** Substitute these definitions into the expression to obtain: \[ \cos \theta \times \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \]3. **Apply identities:** Use the Pythagorean identity to replace \( \cos^2 \theta \) with \( 1 - \sin^2 \theta \), simplifying further: \[ \frac{1 - \sin^2 \theta}{\sin^2 \theta} \]

• Simplification reduces expression complexity, aiding in direct interpretation and problem-solving.
• Key is to continuously substitute known identities to reduce terms progressively.

Being proficient in simplification techniques enhances your ability to analyze and solve a wide range of trigonometric problems efficiently.