Trigonometry (11th Edition)
Trigonometry (11th Edition)
11th Edition
ISBN: 9780134217437
Author: Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher: PEARSON
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**Trigonometric Identity Proof**

**Problem Statement:**

Prove the identity:

\[ \sin(2\theta) = \frac{2\cot(\theta)}{1 + \cot^2(\theta)} \]

**Description:**

This problem requires you to prove a trigonometric identity involving the double angle formula for sine and the cotangent function. The expression on the right side involves cotangent, which is the reciprocal of tangent: \(\cot(\theta) = \frac{1}{\tan(\theta)}\).

The goal is to manipulate one side of the equation to match the other, using trigonometric identities such as:

- Double angle formulas:
  - \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\)
  
- Relationship between sine, cosine, and cotangent:
  - \(\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)}\)

- Pythagorean identity:
  - \(\sin^2(\theta) + \cos^2(\theta) = 1\)

You will likely need to express both sides in terms of sine and cosine, then simplify to demonstrate their equivalence.
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Transcribed Image Text:**Trigonometric Identity Proof** **Problem Statement:** Prove the identity: \[ \sin(2\theta) = \frac{2\cot(\theta)}{1 + \cot^2(\theta)} \] **Description:** This problem requires you to prove a trigonometric identity involving the double angle formula for sine and the cotangent function. The expression on the right side involves cotangent, which is the reciprocal of tangent: \(\cot(\theta) = \frac{1}{\tan(\theta)}\). The goal is to manipulate one side of the equation to match the other, using trigonometric identities such as: - Double angle formulas: - \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) - Relationship between sine, cosine, and cotangent: - \(\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)}\) - Pythagorean identity: - \(\sin^2(\theta) + \cos^2(\theta) = 1\) You will likely need to express both sides in terms of sine and cosine, then simplify to demonstrate their equivalence.
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