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3. If sin(t) = 3 and tan(t) > 0, find the value of each of the six trigonometric functions of t. 4

3. If sin(t) = 3 and tan(t) > 0, find the value of each of the six trigonometric functions of t. 4

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
**Problem 3:** Given that \(\sin(t) = -\frac{3}{4}\) and \(\tan(t) > 0\), find the value of each of the six trigonometric functions of \(t\). **Explanation:** To solve this problem, consider the following steps: 1. **Understanding the Quadrant:** - \(\sin(t) = -\frac{3}{4}\) indicates that sine is negative. - \(\tan(t) > 0\) means that tangent is positive. - Since tangent is positive when both sine and cosine have the same sign, and given sine is negative, both \(\sin(t)\) and \(\cos(t)\) must be negative. Therefore, \(t\) is in the third quadrant. 2. **Finding \(\cos(t)\):** - Use the Pythagorean identity: \(\sin^2(t) + \cos^2(t) = 1\). - \(\left(-\frac{3}{4}\right)^2 + \cos^2(t) = 1\). - \(\frac{9}{16} + \cos^2(t) = 1\). - \(\cos^2(t) = 1 - \frac{9}{16} = \frac{7}{16}\). - \(\cos(t) = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}\) (negative due to the third quadrant). 3. **Finding \(\tan(t)\):** - \(\tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{-\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}}\). 4. **Finding \(\csc(t)\):** - \(\csc(t) = \frac{1}{\sin(t)} = -\frac{4}{3}\). 5. **Finding \(\sec(t)\):** - \(\sec(t) = \frac{1}{\cos(t)} = -\frac{4}{\sqrt{7}}\). 6. **Finding \(\cot(t)\):** - \(\cot(t) = \frac{\cos(t)}{\sin(t)} = \
Transcribed Image Text:**Problem 3:** Given that \(\sin(t) = -\frac{3}{4}\) and \(\tan(t) > 0\), find the value of each of the six trigonometric functions of \(t\). **Explanation:** To solve this problem, consider the following steps: 1. **Understanding the Quadrant:** - \(\sin(t) = -\frac{3}{4}\) indicates that sine is negative. - \(\tan(t) > 0\) means that tangent is positive. - Since tangent is positive when both sine and cosine have the same sign, and given sine is negative, both \(\sin(t)\) and \(\cos(t)\) must be negative. Therefore, \(t\) is in the third quadrant. 2. **Finding \(\cos(t)\):** - Use the Pythagorean identity: \(\sin^2(t) + \cos^2(t) = 1\). - \(\left(-\frac{3}{4}\right)^2 + \cos^2(t) = 1\). - \(\frac{9}{16} + \cos^2(t) = 1\). - \(\cos^2(t) = 1 - \frac{9}{16} = \frac{7}{16}\). - \(\cos(t) = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}\) (negative due to the third quadrant). 3. **Finding \(\tan(t)\):** - \(\tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{-\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}}\). 4. **Finding \(\csc(t)\):** - \(\csc(t) = \frac{1}{\sin(t)} = -\frac{4}{3}\). 5. **Finding \(\sec(t)\):** - \(\sec(t) = \frac{1}{\cos(t)} = -\frac{4}{\sqrt{7}}\). 6. **Finding \(\cot(t)\):** - \(\cot(t) = \frac{\cos(t)}{\sin(t)} = \
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