Show that sin?(0) [1+ cot²(8)] = 1
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Question
![### Mathematical Proof
To prove the given trigonometric identity:
\[
\sin^2(\theta) [1 + \cot^2(\theta)] = 1
\]
#### Proof:
We start with the left-hand side of the equation:
1. We know that \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\). Therefore, \(\cot^2(\theta) = \left(\frac{\cos(\theta)}{\sin(\theta)}\right)^2 = \frac{\cos^2(\theta)}{\sin^2(\theta)}\).
2. Substituting this into the expression, we get:
\[
\sin^2(\theta) \left[1 + \frac{\cos^2(\theta)}{\sin^2(\theta)}\right]
\]
3. Simplifying inside the brackets:
\[
1 + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{\sin^2(\theta)}{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{\sin^2(\theta) + \cos^2(\theta)}{\sin^2(\theta)}
\]
4. Using the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
\[
\frac{1}{\sin^2(\theta)}
\]
5. Now, substituting back into the original expression:
\[
\sin^2(\theta) \times \frac{1}{\sin^2(\theta)} = 1
\]
Hence, the given identity is true:
\[
\boxed{1}
\]
This proof demonstrates that the expression simplifies to 1 through the use of trigonometric identities.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c5734d7-71be-4626-8cb5-227ca6c7fb69%2F0fe5d0e2-6ee1-4a1f-a273-e7db855b63e3%2Fwwenfsq_processed.png&w=3840&q=75)
Transcribed Image Text:### Mathematical Proof
To prove the given trigonometric identity:
\[
\sin^2(\theta) [1 + \cot^2(\theta)] = 1
\]
#### Proof:
We start with the left-hand side of the equation:
1. We know that \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\). Therefore, \(\cot^2(\theta) = \left(\frac{\cos(\theta)}{\sin(\theta)}\right)^2 = \frac{\cos^2(\theta)}{\sin^2(\theta)}\).
2. Substituting this into the expression, we get:
\[
\sin^2(\theta) \left[1 + \frac{\cos^2(\theta)}{\sin^2(\theta)}\right]
\]
3. Simplifying inside the brackets:
\[
1 + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{\sin^2(\theta)}{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{\sin^2(\theta) + \cos^2(\theta)}{\sin^2(\theta)}
\]
4. Using the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
\[
\frac{1}{\sin^2(\theta)}
\]
5. Now, substituting back into the original expression:
\[
\sin^2(\theta) \times \frac{1}{\sin^2(\theta)} = 1
\]
Hence, the given identity is true:
\[
\boxed{1}
\]
This proof demonstrates that the expression simplifies to 1 through the use of trigonometric identities.
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