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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Question
![The problem involves finding \( \tan \theta + \sin \theta \) for the angle \( \theta \) shown in a right triangle.
### Given:
- A right triangle with:
- The opposite side to angle \( \theta \) labeled as \( 2 \).
- The adjacent side to angle \( \theta \) labeled as \( 3 \).
### Required:
- Calculate:
- \( \theta = \underline{\hspace{1cm}} \)
- \( \tan \theta + \sin \theta = \underline{\hspace{1.5cm}} \)
### Diagram Description:
The right triangle has:
- A right angle.
- Angle \( \theta \) is opposite the side of length \( 2 \).
- The side adjacent to \( \theta \) is \( 3 \).
- The hypotenuse is not labeled, implies calculation should use these known sides.
### To Solve:
- Use the Pythagorean theorem to find the hypotenuse:
\[
\text{hypotenuse} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]
- Calculate \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{3} \).
- Calculate \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{13}} \).
- Combine to find \( \tan \theta + \sin \theta \):
\[
\tan \theta + \sin \theta = \frac{2}{3} + \frac{2}{\sqrt{13}}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d456929-ce4c-4399-bf9c-61ec96afc9f0%2Fd93ca690-0e92-47c4-b821-49ae5f2b98d5%2F8ekrmo_processed.png&w=3840&q=75)
Transcribed Image Text:The problem involves finding \( \tan \theta + \sin \theta \) for the angle \( \theta \) shown in a right triangle.
### Given:
- A right triangle with:
- The opposite side to angle \( \theta \) labeled as \( 2 \).
- The adjacent side to angle \( \theta \) labeled as \( 3 \).
### Required:
- Calculate:
- \( \theta = \underline{\hspace{1cm}} \)
- \( \tan \theta + \sin \theta = \underline{\hspace{1.5cm}} \)
### Diagram Description:
The right triangle has:
- A right angle.
- Angle \( \theta \) is opposite the side of length \( 2 \).
- The side adjacent to \( \theta \) is \( 3 \).
- The hypotenuse is not labeled, implies calculation should use these known sides.
### To Solve:
- Use the Pythagorean theorem to find the hypotenuse:
\[
\text{hypotenuse} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]
- Calculate \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{3} \).
- Calculate \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{13}} \).
- Combine to find \( \tan \theta + \sin \theta \):
\[
\tan \theta + \sin \theta = \frac{2}{3} + \frac{2}{\sqrt{13}}
\]
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