177 Give the exact value of Cot
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
![**Question:**
Give the exact value of \( \cot \left( \frac{17\pi}{6} \right) \).
**Explanation:**
This question asks for the exact value of the cotangent function evaluated at the angle \( \frac{17\pi}{6} \). To solve this, follow these steps:
1. Simplify the angle \( \frac{17\pi}{6} \):
Since angles in trigonometry can be coterminal, \( \frac{17\pi}{6} \) can be reduced within one full circle (or \( 2\pi \)).
\[
\frac{17\pi}{6} = \frac{17}{6} \pi
\]
Subtract \( 2\pi \) (since one full circle is \( 2\pi \)):
\[
\frac{17}{6}\pi - 2\pi = \frac{17\pi - 12\pi}{6} = \frac{5\pi}{6}
\]
This means that \( \frac{17\pi}{6} \) is coterminal with \( \frac{5\pi}{6} \).
2. Evaluate \(\cot \left( \frac{5\pi}{6} \right) \):
Cotangent is the reciprocal of tangent:
\[
\cot \theta = \frac{1}{\tan \theta}
\]
For \(\theta = \frac{5\pi}{6} \):
\[
\cot \left( \frac{5\pi}{6} \right) = \frac{1}{\tan \left( \frac{5\pi}{6} \right)}
\]
The tangent function of \( \frac{5\pi}{6} \) can be evaluated considering it is in the second quadrant:
Tangent in the second quadrant:
\[
\tan \left( \frac{5\pi}{6} \right) = - \tan \left( \pi - \frac{\pi}{6} \right) = - \tan \left( \frac{\pi}{6} \right) = - \frac{1}{\sqrt{3}}
\]
Hence:
\[
\cot \left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21d92c2f-6485-4b0b-a513-40194c3c2abd%2F54b4db84-fb38-4a76-8122-0f05c249258a%2Fj758qkj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
Give the exact value of \( \cot \left( \frac{17\pi}{6} \right) \).
**Explanation:**
This question asks for the exact value of the cotangent function evaluated at the angle \( \frac{17\pi}{6} \). To solve this, follow these steps:
1. Simplify the angle \( \frac{17\pi}{6} \):
Since angles in trigonometry can be coterminal, \( \frac{17\pi}{6} \) can be reduced within one full circle (or \( 2\pi \)).
\[
\frac{17\pi}{6} = \frac{17}{6} \pi
\]
Subtract \( 2\pi \) (since one full circle is \( 2\pi \)):
\[
\frac{17}{6}\pi - 2\pi = \frac{17\pi - 12\pi}{6} = \frac{5\pi}{6}
\]
This means that \( \frac{17\pi}{6} \) is coterminal with \( \frac{5\pi}{6} \).
2. Evaluate \(\cot \left( \frac{5\pi}{6} \right) \):
Cotangent is the reciprocal of tangent:
\[
\cot \theta = \frac{1}{\tan \theta}
\]
For \(\theta = \frac{5\pi}{6} \):
\[
\cot \left( \frac{5\pi}{6} \right) = \frac{1}{\tan \left( \frac{5\pi}{6} \right)}
\]
The tangent function of \( \frac{5\pi}{6} \) can be evaluated considering it is in the second quadrant:
Tangent in the second quadrant:
\[
\tan \left( \frac{5\pi}{6} \right) = - \tan \left( \pi - \frac{\pi}{6} \right) = - \tan \left( \frac{\pi}{6} \right) = - \frac{1}{\sqrt{3}}
\]
Hence:
\[
\cot \left(
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