-4 sin 0 = Find cose jif 5 and 0 terminates in QIV.
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![Transcription for Educational Website:
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Watch the signs of your answer.
Find cos(θ) if sin(θ) = -4/5 and θ terminates in QIV.
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Here, we are given that sin(θ) = -4/5 and θ is in the fourth quadrant (QIV). The task is to find the value of cos(θ).
In the fourth quadrant, the cosine function is positive. To find cos(θ), we can use the Pythagorean identity:
\[ \sin^2(θ) + \cos^2(θ) = 1 \]
Given:
\[ \sin(θ) = -4/5 \]
First, square the given sine value:
\[ (-4/5)^2 = 16/25 \]
Now, substitute this into the Pythagorean identity:
\[ (16/25) + \cos^2(θ) = 1 \]
Solving for \(\cos^2(θ)\):
\[ \cos^2(θ) = 1 - 16/25 \]
\[ \cos^2(θ) = 25/25 - 16/25 \]
\[ \cos^2(θ) = 9/25 \]
Taking the square root of both sides, we get:
\[ \cos(θ) = ±√(9/25) \]
\[ \cos(θ) = ±3/5 \]
Since θ is in the fourth quadrant where cosine is positive:
\[ \cos(θ) = 3/5 \]
Therefore, the value of \(\cos(θ)\) is \( 3/5 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F242c3500-5eba-4bd1-9201-b2fb9263f121%2Fa97607e8-f212-4f9a-95e5-8c7d25b62848%2Fi5ggdij.jpeg&w=3840&q=75)
Transcribed Image Text:Transcription for Educational Website:
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Watch the signs of your answer.
Find cos(θ) if sin(θ) = -4/5 and θ terminates in QIV.
---
Here, we are given that sin(θ) = -4/5 and θ is in the fourth quadrant (QIV). The task is to find the value of cos(θ).
In the fourth quadrant, the cosine function is positive. To find cos(θ), we can use the Pythagorean identity:
\[ \sin^2(θ) + \cos^2(θ) = 1 \]
Given:
\[ \sin(θ) = -4/5 \]
First, square the given sine value:
\[ (-4/5)^2 = 16/25 \]
Now, substitute this into the Pythagorean identity:
\[ (16/25) + \cos^2(θ) = 1 \]
Solving for \(\cos^2(θ)\):
\[ \cos^2(θ) = 1 - 16/25 \]
\[ \cos^2(θ) = 25/25 - 16/25 \]
\[ \cos^2(θ) = 9/25 \]
Taking the square root of both sides, we get:
\[ \cos(θ) = ±√(9/25) \]
\[ \cos(θ) = ±3/5 \]
Since θ is in the fourth quadrant where cosine is positive:
\[ \cos(θ) = 3/5 \]
Therefore, the value of \(\cos(θ)\) is \( 3/5 \).
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