Find the exact value for the expression under the given conditions. 5 5 cos (a +ß), sin a = - for a in Quadrant III and cos ß=- 13 12 cos (a + B)= X for ß in Quadrant II.
Find the exact value for the expression under the given conditions. 5 5 cos (a +ß), sin a = - for a in Quadrant III and cos ß=- 13 12 cos (a + B)= X for ß in Quadrant II.
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°.
Related questions
Question
![**Problem Statement:**
Find the exact value for the expression under the given conditions.
Given:
\[
\cos (\alpha + \beta), \quad \sin \alpha = -\frac{5}{13} \quad \text{for } \alpha \text{ in Quadrant III} \quad \text{and}\quad \cos \beta = -\frac{5}{12} \quad \text{for } \beta \text{ in Quadrant II}.
\]
**Calculation Area:**
\[
\cos (\alpha + \beta) = \text{ (input box)}
\]
There is an interactive area provided to input and verify your answer. The interface allows you to enter mathematical symbols and expressions.
**Solution Steps Overview:**
To solve for \(\cos(\alpha + \beta)\), we can use the angle addition formula for cosine:
\[
\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
\]
Given values are:
\[
\sin \alpha = -\frac{5}{13} \quad \text{(for } \alpha \text{ in Quadrant III)}
\]
\[
\cos \beta = -\frac{5}{12} \quad \text{(for } \beta \text{ in Quadrant II)}
\]
1. First find \(\cos \alpha\):
In Quadrant III, both sine and cosine are negative. Using the Pythagorean identity:
\[
\sin^2 \alpha + \cos^2 \alpha = 1
\]
\[
\left(-\frac{5}{13}\right)^2 + \cos^2 \alpha = 1
\]
\[
\frac{25}{169} + \cos^2 \alpha = 1
\]
\[
\cos^2 \alpha = 1 - \frac{25}{169}
\]
\[
\cos^2 \alpha = \frac{144}{169}
\]
Since \(\alpha\) is in Quadrant III, \(\cos \alpha\) is negative:
\[
\cos \alpha = -\frac{12}{13}
\]
2. Find \(\sin \beta\):
In Quadrant II, sine is positive and cosine is negative. Using the Pythagorean identity:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe92e694a-8dce-4f5c-a0b3-7db66cedb1e5%2Ff5085e43-5d29-44e6-948c-26a99e41b8fb%2Fafwsxhp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the exact value for the expression under the given conditions.
Given:
\[
\cos (\alpha + \beta), \quad \sin \alpha = -\frac{5}{13} \quad \text{for } \alpha \text{ in Quadrant III} \quad \text{and}\quad \cos \beta = -\frac{5}{12} \quad \text{for } \beta \text{ in Quadrant II}.
\]
**Calculation Area:**
\[
\cos (\alpha + \beta) = \text{ (input box)}
\]
There is an interactive area provided to input and verify your answer. The interface allows you to enter mathematical symbols and expressions.
**Solution Steps Overview:**
To solve for \(\cos(\alpha + \beta)\), we can use the angle addition formula for cosine:
\[
\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
\]
Given values are:
\[
\sin \alpha = -\frac{5}{13} \quad \text{(for } \alpha \text{ in Quadrant III)}
\]
\[
\cos \beta = -\frac{5}{12} \quad \text{(for } \beta \text{ in Quadrant II)}
\]
1. First find \(\cos \alpha\):
In Quadrant III, both sine and cosine are negative. Using the Pythagorean identity:
\[
\sin^2 \alpha + \cos^2 \alpha = 1
\]
\[
\left(-\frac{5}{13}\right)^2 + \cos^2 \alpha = 1
\]
\[
\frac{25}{169} + \cos^2 \alpha = 1
\]
\[
\cos^2 \alpha = 1 - \frac{25}{169}
\]
\[
\cos^2 \alpha = \frac{144}{169}
\]
Since \(\alpha\) is in Quadrant III, \(\cos \alpha\) is negative:
\[
\cos \alpha = -\frac{12}{13}
\]
2. Find \(\sin \beta\):
In Quadrant II, sine is positive and cosine is negative. Using the Pythagorean identity:
\[
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