10. If 0° 0 0° ≤ 0 ≤ 3 ne = , find cos(20) 5 ≤ 90° and sine
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
![**Trigonometry Problem**
**Problem:**
10. If \(0^\circ \leq \theta \leq 90^\circ\) and \(\sin\theta = \frac{3}{5}\), find \(\cos(2\theta)\).
**Solution:**
Given that \(\sin\theta = \frac{3}{5}\), we can use the Pythagorean identity to find \(\cos\theta\). The Pythagorean identity states:
\[
\sin^2\theta + \cos^2\theta = 1
\]
Substituting the given value of \(\sin\theta\):
\[
\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1
\]
Simplifying:
\[
\frac{9}{25} + \cos^2\theta = 1
\]
Next, solve for \(\cos^2\theta\):
\[
\cos^2\theta = 1 - \frac{9}{25}
\]
\[
\cos^2\theta = \frac{25}{25} - \frac{9}{25}
\]
\[
\cos^2\theta = \frac{16}{25}
\]
Take the square root of both sides to find \(\cos\theta\):
\[
\cos\theta = \pm\frac{4}{5}
\]
Since \(0^\circ \leq \theta \leq 90^\circ\), \(\cos\theta\) is positive, so:
\[
\cos\theta = \frac{4}{5}
\]
Now, we need to find \(\cos(2\theta)\). Using the double-angle formula for cosine:
\[
\cos(2\theta) = 2\cos^2\theta - 1
\]
Substitute \(\cos\theta = \frac{4}{5}\):
\[
\cos(2\theta) = 2 \left(\frac{4}{5}\right)^2 - 1
\]
\[
\cos(2\theta) = 2 \left(\frac{16}{25}\right) - 1
\]
\[
\cos(2\theta) = \frac{32}{25} - 1
\]
\[
\cos(2\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F050b7896-96e1-4a53-ae28-da4cd34956eb%2Fe049392c-4c27-4dfc-ab58-00e0ad0f02d3%2Fcj3f22xp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Trigonometry Problem**
**Problem:**
10. If \(0^\circ \leq \theta \leq 90^\circ\) and \(\sin\theta = \frac{3}{5}\), find \(\cos(2\theta)\).
**Solution:**
Given that \(\sin\theta = \frac{3}{5}\), we can use the Pythagorean identity to find \(\cos\theta\). The Pythagorean identity states:
\[
\sin^2\theta + \cos^2\theta = 1
\]
Substituting the given value of \(\sin\theta\):
\[
\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1
\]
Simplifying:
\[
\frac{9}{25} + \cos^2\theta = 1
\]
Next, solve for \(\cos^2\theta\):
\[
\cos^2\theta = 1 - \frac{9}{25}
\]
\[
\cos^2\theta = \frac{25}{25} - \frac{9}{25}
\]
\[
\cos^2\theta = \frac{16}{25}
\]
Take the square root of both sides to find \(\cos\theta\):
\[
\cos\theta = \pm\frac{4}{5}
\]
Since \(0^\circ \leq \theta \leq 90^\circ\), \(\cos\theta\) is positive, so:
\[
\cos\theta = \frac{4}{5}
\]
Now, we need to find \(\cos(2\theta)\). Using the double-angle formula for cosine:
\[
\cos(2\theta) = 2\cos^2\theta - 1
\]
Substitute \(\cos\theta = \frac{4}{5}\):
\[
\cos(2\theta) = 2 \left(\frac{4}{5}\right)^2 - 1
\]
\[
\cos(2\theta) = 2 \left(\frac{16}{25}\right) - 1
\]
\[
\cos(2\theta) = \frac{32}{25} - 1
\]
\[
\cos(2\
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