21 Cos A 29 20 Suppose A is an acute angle, and sin A 29 Find sin 2A and cos 2A. sin 2A = cos 2A

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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**Example Problem: Find sin(2A) and cos(2A)**

Suppose \( A \) is an acute angle, and 

\[ \sin A = \frac{21}{29} \]
\[ \cos A = \frac{20}{29} \]

Find \( \sin 2A \) and \( \cos 2A \).

\[ \sin 2A = \]
\[ \cos 2A = \]

### Solution:
To find \( \sin 2A \) and \( \cos 2A \), we can use the following double-angle identities:

1. \( \sin 2A = 2 \sin A \cos A \)
2. \( \cos 2A = \cos^2 A - \sin^2 A \)

Using the given values:
\[ \sin A = \frac{21}{29} \]
\[ \cos A = \frac{20}{29} \]

First, calculate \( \sin 2A \):

\[ \sin 2A = 2 \sin A \cos A \]
\[ \sin 2A = 2 \left(\frac{21}{29}\right) \left(\frac{20}{29}\right) \]
\[ \sin 2A = 2 \cdot \frac{420}{841} \]
\[ \sin 2A = \frac{840}{841} \]

Now, calculate \( \cos 2A \):

\[ \cos 2A = \cos^2 A - \sin^2 A \]
\[ \cos 2A = \left(\frac{20}{29}\right)^2 - \left(\frac{21}{29}\right)^2 \]
\[ \cos 2A = \frac{400}{841} - \frac{441}{841} \]
\[ \cos 2A = \frac{400 - 441}{841} \]
\[ \cos 2A = \frac{-41}{841} \]

Putting these results in the answer fields:

\[ \sin 2A = \frac{840}{841} \]
\[ \cos 2A = \frac{-41}{841} \]
Transcribed Image Text:**Example Problem: Find sin(2A) and cos(2A)** Suppose \( A \) is an acute angle, and \[ \sin A = \frac{21}{29} \] \[ \cos A = \frac{20}{29} \] Find \( \sin 2A \) and \( \cos 2A \). \[ \sin 2A = \] \[ \cos 2A = \] ### Solution: To find \( \sin 2A \) and \( \cos 2A \), we can use the following double-angle identities: 1. \( \sin 2A = 2 \sin A \cos A \) 2. \( \cos 2A = \cos^2 A - \sin^2 A \) Using the given values: \[ \sin A = \frac{21}{29} \] \[ \cos A = \frac{20}{29} \] First, calculate \( \sin 2A \): \[ \sin 2A = 2 \sin A \cos A \] \[ \sin 2A = 2 \left(\frac{21}{29}\right) \left(\frac{20}{29}\right) \] \[ \sin 2A = 2 \cdot \frac{420}{841} \] \[ \sin 2A = \frac{840}{841} \] Now, calculate \( \cos 2A \): \[ \cos 2A = \cos^2 A - \sin^2 A \] \[ \cos 2A = \left(\frac{20}{29}\right)^2 - \left(\frac{21}{29}\right)^2 \] \[ \cos 2A = \frac{400}{841} - \frac{441}{841} \] \[ \cos 2A = \frac{400 - 441}{841} \] \[ \cos 2A = \frac{-41}{841} \] Putting these results in the answer fields: \[ \sin 2A = \frac{840}{841} \] \[ \cos 2A = \frac{-41}{841} \]
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