### Trigonometric Identities ProblemIf \( A \) 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.### ExplanationTo 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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Answered: 1 If A is an angle in QIII such that…

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}}

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