Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Trigonometry Problem**
**Problem Statement:**
7. Find the value of \( \sin (\alpha - \beta) \) if \( \tan \alpha = \frac{4}{3} \), \( \cot \beta = \frac{5}{12} \), with the conditions \( 0^\circ < \alpha < 90^\circ \) and \( 0^\circ < \beta < 90^\circ \).
**Solution Approach:**
To find \( \sin (\alpha - \beta) \), we need to use the trigonometric identity:
\[
\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta
\]
Given:
- \( \tan \alpha = \frac{4}{3} \) implies a right triangle where the opposite side is 4, the adjacent side is 3, and the hypotenuse is \( \sqrt{4^2 + 3^2} = 5 \). Therefore:
- \( \sin \alpha = \frac{4}{5} \)
- \( \cos \alpha = \frac{3}{5} \)
- \( \cot \beta = \frac{5}{12} \) implies a right triangle where the adjacent side is 5, the opposite side is 12, and the hypotenuse is \( \sqrt{5^2 + 12^2} = 13 \). Therefore:
- \( \sin \beta = \frac{12}{13} \)
- \( \cos \beta = \frac{5}{13} \)
Plugging these into the identity:
\[
\sin (\alpha - \beta) = \left( \frac{4}{5} \times \frac{5}{13} \right) - \left( \frac{3}{5} \times \frac{12}{13} \right)
\]
Simplifying:
\[
= \left( \frac{20}{65} \right) - \left( \frac{36}{65} \right)
= \frac{20 - 36}{65}
= \frac{-16}{65}
\]
Hence, the value of \( \sin(\alpha - \beta) \) is \( \frac{-16}{65} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb805b29b-1703-42be-83a9-c544ba7b8f1b%2Ff032558c-3c8f-41b2-8418-01f20b511b63%2Fixwf82u_processed.png&w=3840&q=75)
Transcribed Image Text:**Trigonometry Problem**
**Problem Statement:**
7. Find the value of \( \sin (\alpha - \beta) \) if \( \tan \alpha = \frac{4}{3} \), \( \cot \beta = \frac{5}{12} \), with the conditions \( 0^\circ < \alpha < 90^\circ \) and \( 0^\circ < \beta < 90^\circ \).
**Solution Approach:**
To find \( \sin (\alpha - \beta) \), we need to use the trigonometric identity:
\[
\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta
\]
Given:
- \( \tan \alpha = \frac{4}{3} \) implies a right triangle where the opposite side is 4, the adjacent side is 3, and the hypotenuse is \( \sqrt{4^2 + 3^2} = 5 \). Therefore:
- \( \sin \alpha = \frac{4}{5} \)
- \( \cos \alpha = \frac{3}{5} \)
- \( \cot \beta = \frac{5}{12} \) implies a right triangle where the adjacent side is 5, the opposite side is 12, and the hypotenuse is \( \sqrt{5^2 + 12^2} = 13 \). Therefore:
- \( \sin \beta = \frac{12}{13} \)
- \( \cos \beta = \frac{5}{13} \)
Plugging these into the identity:
\[
\sin (\alpha - \beta) = \left( \frac{4}{5} \times \frac{5}{13} \right) - \left( \frac{3}{5} \times \frac{12}{13} \right)
\]
Simplifying:
\[
= \left( \frac{20}{65} \right) - \left( \frac{36}{65} \right)
= \frac{20 - 36}{65}
= \frac{-16}{65}
\]
Hence, the value of \( \sin(\alpha - \beta) \) is \( \frac{-16}{65} \).
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