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V17 Find all possible values of cos(0). 9. Let sin(0)

V17 Find all possible values of cos(0). 9. Let sin(0)

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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**Problem Statement:** Let \(\sin(\theta) = \frac{\sqrt{17}}{9}\). Find all possible values of \(\cos(\theta)\). **Solution:** To determine the possible values of \(\cos(\theta)\), we can use the fundamental Pythagorean identity which is: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Given: \[ \sin(\theta) = \frac{\sqrt{17}}{9} \] First, square both sides to find \(\sin^2(\theta)\): \[ \sin^2(\theta) = \left(\frac{\sqrt{17}}{9}\right)^2 = \frac{17}{81} \] Substitute this back into the Pythagorean identity: \[ \frac{17}{81} + \cos^2(\theta) = 1 \] Solve for \(\cos^2(\theta)\): \[ \cos^2(\theta) = 1 - \frac{17}{81} = \frac{81}{81} - \frac{17}{81} = \frac{64}{81} \] Therefore, \(\cos(\theta)\) can be either the positive or negative square root of \(\frac{64}{81}\): \[ \cos(\theta) = \pm \frac{8}{9} \] **Answer:** \[ \cos(\theta) = \frac{8}{9} \quad \text{or} \quad \cos(\theta) = -\frac{8}{9} \] These are the possible values of \(\cos(\theta)\) given that \(\sin(\theta) = \frac{\sqrt{17}}{9}\).
Transcribed Image Text:**Problem Statement:** Let \(\sin(\theta) = \frac{\sqrt{17}}{9}\). Find all possible values of \(\cos(\theta)\). **Solution:** To determine the possible values of \(\cos(\theta)\), we can use the fundamental Pythagorean identity which is: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Given: \[ \sin(\theta) = \frac{\sqrt{17}}{9} \] First, square both sides to find \(\sin^2(\theta)\): \[ \sin^2(\theta) = \left(\frac{\sqrt{17}}{9}\right)^2 = \frac{17}{81} \] Substitute this back into the Pythagorean identity: \[ \frac{17}{81} + \cos^2(\theta) = 1 \] Solve for \(\cos^2(\theta)\): \[ \cos^2(\theta) = 1 - \frac{17}{81} = \frac{81}{81} - \frac{17}{81} = \frac{64}{81} \] Therefore, \(\cos(\theta)\) can be either the positive or negative square root of \(\frac{64}{81}\): \[ \cos(\theta) = \pm \frac{8}{9} \] **Answer:** \[ \cos(\theta) = \frac{8}{9} \quad \text{or} \quad \cos(\theta) = -\frac{8}{9} \] These are the possible values of \(\cos(\theta)\) given that \(\sin(\theta) = \frac{\sqrt{17}}{9}\).
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