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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![**Problem Statement**
Find \(\sin t\) and \(\cos t\) for the given value of \(t\).
\[ t = 480^\circ \]
**Solutions**
\[\sin t = \_\_\_\_\_\_\_\_\_\_\]
\[\cos t = \_\_\_\_\_\_\_\_\_\_\]
**Explanation**
To find \(\sin t\) and \(\cos t\) for an angle of \(480^\circ\), first reduce the angle as follows:
1. Subtract \(360^\circ\) from the given angle to find its equivalent within the standard \(0^\circ\) to \(360^\circ\) range.
2. \(480^\circ - 360^\circ = 120^\circ\).
Now, use \(120^\circ\) to find the corresponding sine and cosine values.
**Trigonometric Values**
- \(\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}\)
- \(\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}\)
Therefore:
\[ \sin 480^\circ = \frac{\sqrt{3}}{2} \]
\[ \cos 480^\circ = -\frac{1}{2} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a328885-7641-4d6d-85af-435d99e02dcf%2F04562640-8228-4170-8746-6ae836fd5002%2Fhzkcrnf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find \(\sin t\) and \(\cos t\) for the given value of \(t\).
\[ t = 480^\circ \]
**Solutions**
\[\sin t = \_\_\_\_\_\_\_\_\_\_\]
\[\cos t = \_\_\_\_\_\_\_\_\_\_\]
**Explanation**
To find \(\sin t\) and \(\cos t\) for an angle of \(480^\circ\), first reduce the angle as follows:
1. Subtract \(360^\circ\) from the given angle to find its equivalent within the standard \(0^\circ\) to \(360^\circ\) range.
2. \(480^\circ - 360^\circ = 120^\circ\).
Now, use \(120^\circ\) to find the corresponding sine and cosine values.
**Trigonometric Values**
- \(\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}\)
- \(\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}\)
Therefore:
\[ \sin 480^\circ = \frac{\sqrt{3}}{2} \]
\[ \cos 480^\circ = -\frac{1}{2} \]
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