1 If A is an angle in QIII such that sin A = -- and B is an angle in QIV such that sin B to find cos A = cos B= sin (A + B) = cos(A + B) = 2 7 use identities
1 If A is an angle in QIII such that sin A = -- and B is an angle in QIV such that sin B to find cos A = cos B= sin (A + B) = cos(A + B) = 2 7 use identities
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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Question
![### Trigonometric Identities Problem
If \( A \) is an angle in Quadrant III such that \( \sin A = -\frac{1}{4} \) and \( B \) is an angle in Quadrant IV such that \( \sin B = -\frac{2}{7} \), use trigonometric identities to find:
1. \(\cos A\)
**Solution:** Box for filling in the answer.
2. \(\cos B\)
**Solution:** Box for filling in the answer.
3. \(\sin(A + B)\)
**Solution:** Box for filling in the answer.
4. \(\cos(A + B)\)
**Solution:** Box for filling in the answer.
### Explanation
To solve for these values, we will use various trigonometric identities and properties based on the given conditions.
#### 1. Finding \(\cos A\):
We know that in Quadrant III, the cosine of an angle is negative. We use the Pythagorean identity to find \(\cos A\):
\[
\sin^2 A + \cos^2 A = 1 \implies \left( -\frac{1}{4} \right)^2 + \cos^2 A = 1 \implies \frac{1}{16} + \cos^2 A = 1 \implies \cos^2 A = 1 - \frac{1}{16} \implies \cos^2 A = \frac{15}{16} \implies \cos A = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}
\]
#### 2. Finding \(\cos B\):
Similarly, we know that in Quadrant IV, the cosine of an angle is positive. Using the Pythagorean identity:
\[
\sin^2 B + \cos^2 B = 1 \implies \left( -\frac{2}{7} \right)^2 + \cos^2 B = 1 \implies \frac{4}{49} + \cos^2 B = 1 \implies \cos^2 B = 1 - \frac{4}{49} \implies \cos^2 B = \frac{45}{49} \implies \cos B = \sqrt{\frac{45}{49}}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b559ab3-5119-482d-94d5-a98a530a62e8%2F50925228-c644-421a-adb3-ba9974389113%2F0hxawko_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Trigonometric Identities Problem
If \( A \) is an angle in Quadrant III such that \( \sin A = -\frac{1}{4} \) and \( B \) is an angle in Quadrant IV such that \( \sin B = -\frac{2}{7} \), use trigonometric identities to find:
1. \(\cos A\)
**Solution:** Box for filling in the answer.
2. \(\cos B\)
**Solution:** Box for filling in the answer.
3. \(\sin(A + B)\)
**Solution:** Box for filling in the answer.
4. \(\cos(A + B)\)
**Solution:** Box for filling in the answer.
### Explanation
To solve for these values, we will use various trigonometric identities and properties based on the given conditions.
#### 1. Finding \(\cos A\):
We know that in Quadrant III, the cosine of an angle is negative. We use the Pythagorean identity to find \(\cos A\):
\[
\sin^2 A + \cos^2 A = 1 \implies \left( -\frac{1}{4} \right)^2 + \cos^2 A = 1 \implies \frac{1}{16} + \cos^2 A = 1 \implies \cos^2 A = 1 - \frac{1}{16} \implies \cos^2 A = \frac{15}{16} \implies \cos A = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}
\]
#### 2. Finding \(\cos B\):
Similarly, we know that in Quadrant IV, the cosine of an angle is positive. Using the Pythagorean identity:
\[
\sin^2 B + \cos^2 B = 1 \implies \left( -\frac{2}{7} \right)^2 + \cos^2 B = 1 \implies \frac{4}{49} + \cos^2 B = 1 \implies \cos^2 B = 1 - \frac{4}{49} \implies \cos^2 B = \frac{45}{49} \implies \cos B = \sqrt{\frac{45}{49}}
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