Find the exact value of cos a, given that sin a = and a is in quadrant I. 11

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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Title: Cosine Value Calculation

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**Problem Description:**

Find the exact value of \( \cos \, \alpha \) given that \( \sin \, \alpha = \frac{7}{11} \) and \( \alpha \) is in quadrant I.

\[ \cos \, \alpha = \boxed{} \]
*(Simplify your answer. Type an exact answer, using radicals as needed.)*

**Instructions:**

1. Enter your answer in the answer box provided and then click "Check Answer."

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**Explanation:**

To find the cosine of the angle \( \alpha \), we use the relationship between the sine and cosine in a right triangle as well as trigonometric identities.

Given:
\[ \sin \, \alpha = \frac{7}{11} \]
Since \( \alpha \) is in quadrant I, both sine and cosine values will be positive.

We know the fundamental trigonometric identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

First, we square the given sine value:
\[ \left(\frac{7}{11}\right)^2 = \frac{49}{121} \]

Then, we use the identity to solve for \( \cos^2 \alpha \):
\[ \cos^2 \alpha = 1 - \sin^2 \alpha \]
\[ \cos^2 \alpha = 1 - \frac{49}{121} \]
\[ \cos^2 \alpha = \frac{121}{121} - \frac{49}{121} \]
\[ \cos^2 \alpha = \frac{72}{121} \]

Next, we take the square root to find \( \cos \, \alpha \):
\[ \cos \alpha = \sqrt{\frac{72}{121}} \]
\[ \cos \alpha = \frac{\sqrt{72}}{11} \]
\[ \cos \alpha = \frac{6\sqrt{2}}{11} \]

Therefore, the exact value of \( \cos \alpha \) is:
\[ \cos \alpha = \frac{6\sqrt{2}}{11} \]

*[Please input your answer in the box provided and check to confirm it's correct.]*
Transcribed Image Text:Title: Cosine Value Calculation --- **Problem Description:** Find the exact value of \( \cos \, \alpha \) given that \( \sin \, \alpha = \frac{7}{11} \) and \( \alpha \) is in quadrant I. \[ \cos \, \alpha = \boxed{} \] *(Simplify your answer. Type an exact answer, using radicals as needed.)* **Instructions:** 1. Enter your answer in the answer box provided and then click "Check Answer." --- **Explanation:** To find the cosine of the angle \( \alpha \), we use the relationship between the sine and cosine in a right triangle as well as trigonometric identities. Given: \[ \sin \, \alpha = \frac{7}{11} \] Since \( \alpha \) is in quadrant I, both sine and cosine values will be positive. We know the fundamental trigonometric identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] First, we square the given sine value: \[ \left(\frac{7}{11}\right)^2 = \frac{49}{121} \] Then, we use the identity to solve for \( \cos^2 \alpha \): \[ \cos^2 \alpha = 1 - \sin^2 \alpha \] \[ \cos^2 \alpha = 1 - \frac{49}{121} \] \[ \cos^2 \alpha = \frac{121}{121} - \frac{49}{121} \] \[ \cos^2 \alpha = \frac{72}{121} \] Next, we take the square root to find \( \cos \, \alpha \): \[ \cos \alpha = \sqrt{\frac{72}{121}} \] \[ \cos \alpha = \frac{\sqrt{72}}{11} \] \[ \cos \alpha = \frac{6\sqrt{2}}{11} \] Therefore, the exact value of \( \cos \alpha \) is: \[ \cos \alpha = \frac{6\sqrt{2}}{11} \] *[Please input your answer in the box provided and check to confirm it's correct.]*
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