Chapter 7: Problem 109
If \(\sin \theta=\frac{1}{5},\) find \(\csc (\theta+\pi)\)
Short Answer
Expert verified
\text{csc}(\theta + \pi) = -5
Step by step solution
01
Identify the Given Information
Given \(\text{sin}(\theta) = \frac{1}{5} \), identify that sin is provided for angle \( \theta \).
02
Recall the Definition of Cosecant
Recall that \( \text{csc}(x) = \frac{1}{\text{sin}(x)} \).
03
Use the Periodicity of Trigonometric Functions
Remember that \( \text{csc}(\theta + \pi) \) relates to the periodic properties of trigonometric functions. Specifically, \( \text{sin}(\theta + \pi) = -\text{sin}(\theta) \).
04
Substitute the Given Value
Substitute the given \( \text{sin}(\theta) = \frac{1}{5} \) into the equation: \( \text{sin}(\theta + \pi) = -\frac{1}{5} \).
05
Calculate the Required Cosecant
Using the definition of cosecant, compute \( \text{csc}(\theta + \pi) = \frac{1}{\text{sin}(\theta + \pi)} = \frac{1}{-\frac{1}{5}} = -5 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
cosecant
The cosecant function, abbreviated as \(\text{csc}(x)\), is the reciprocal of the sine function. Remember, the sine function gives the ratio of the opposite side to the hypotenuse in a right triangle. Therefore: \(\text{csc}(x) = \frac{1}{\text{sin}(x)}\).
So, if \(\text{sin}(x)\) is known, you can find \(\text{csc}(x)\) by simply taking its reciprocal.
This relationship allows us to solve problems involving sine and cosecant easily.
In the given problem, knowing that \(\text{sin}(\theta) = \frac{1}{5}\), we find the cosecant directly as \(\text{csc}(\theta) = \frac{1}{\frac{1}{5}} = 5\).
So, if \(\text{sin}(x)\) is known, you can find \(\text{csc}(x)\) by simply taking its reciprocal.
This relationship allows us to solve problems involving sine and cosecant easily.
In the given problem, knowing that \(\text{sin}(\theta) = \frac{1}{5}\), we find the cosecant directly as \(\text{csc}(\theta) = \frac{1}{\frac{1}{5}} = 5\).
trigonometric periodicity
Trigonometric functions like sine and cosecant have periodic properties. This means their values repeat at regular intervals. For sine and cosecant, the period is \(\text{2}\text{Ï€}\).
This periodicity helps in simplifying expressions involving shifted angles. For instance, the sine function has a property:
This relation tells us that if you shift the angle \(\theta\) by \(\text{Ï€}\), the sine value simply becomes the negative of what it was.
Using this property in the problem, \(\text{sin}(\theta + \text{Ï€}) = - \frac{1}{5}\).
Periodic properties make trigonometric problem-solving more straightforward, especially when dealing with angles beyond the typical range.
This periodicity helps in simplifying expressions involving shifted angles. For instance, the sine function has a property:
- \(\text{sin}(\theta + \text{Ï€}) = -\text{sin}(\theta)\)
This relation tells us that if you shift the angle \(\theta\) by \(\text{Ï€}\), the sine value simply becomes the negative of what it was.
Using this property in the problem, \(\text{sin}(\theta + \text{Ï€}) = - \frac{1}{5}\).
Periodic properties make trigonometric problem-solving more straightforward, especially when dealing with angles beyond the typical range.
sine function
The sine function is fundamental in trigonometry. It describes a smooth, wave-like motion. In a right triangle, \(\text{sin}(x)\) gives the ratio of the length of the opposite side to the hypotenuse. The sine function has a range of [-1, 1] and is periodic with a period of \(\text{2}\text{Ï€}\).
If \(\text{sin}(x)\) is known, other trigonometric functions can also be determined.
In the problem, we have \(\text{sin}(\theta) = \frac{1}{5}\). Since the sine function is periodic, \(\text{sin}(\theta + \text{Ï€})\) can be calculated using the property:
Substituting the known value: \(\text{sin}(\theta + \text{Ï€}) = - \frac{1}{5} \).
Understanding the sine function helps in various mathematical and real-world applications, including waves, oscillations, and circular motion.
If \(\text{sin}(x)\) is known, other trigonometric functions can also be determined.
In the problem, we have \(\text{sin}(\theta) = \frac{1}{5}\). Since the sine function is periodic, \(\text{sin}(\theta + \text{Ï€})\) can be calculated using the property:
- \(\text{sin}(\theta + \text{Ï€}) = - \text{sin}(\theta) \)
Substituting the known value: \(\text{sin}(\theta + \text{Ï€}) = - \frac{1}{5} \).
Understanding the sine function helps in various mathematical and real-world applications, including waves, oscillations, and circular motion.