1 3. If sin x = and is in quadrant I, use the double-angle identities to find sin (2x), cos (2x), and tan (2x). (Hint: find cos x and tan x first.)
1 3. If sin x = and is in quadrant I, use the double-angle identities to find sin (2x), cos (2x), and tan (2x). (Hint: find cos x and tan x first.)
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
![**Problem 3:**
Given that \(\sin x = \frac{1}{8}\) and \(x\) is in quadrant I, use the double-angle identities to find \(\sin(2x)\), \(\cos(2x)\), and \(\tan(2x)\).
*(Hint: find \(\cos x\) and \(\tan x\) first.)*
**Solution Approach:**
1. **Use Pythagorean Identity:**
\[
\sin^2 x + \cos^2 x = 1
\]
Substitute \(\sin x = \frac{1}{8}\):
\[
\left(\frac{1}{8}\right)^2 + \cos^2 x = 1 \implies \frac{1}{64} + \cos^2 x = 1
\]
\[
\cos^2 x = 1 - \frac{1}{64} = \frac{63}{64}
\]
Since \(x\) is in quadrant I:
\[
\cos x = \frac{\sqrt{63}}{8}
\]
2. **Find \(\sin(2x)\) using double-angle identity:**
\[
\sin(2x) = 2 \sin x \cos x
\]
\[
\sin(2x) = 2 \left(\frac{1}{8}\right) \left(\frac{\sqrt{63}}{8}\right) = \frac{\sqrt{63}}{32}
\]
3. **Find \(\cos(2x)\) using double-angle identity:**
\[
\cos(2x) = \cos^2 x - \sin^2 x
\]
\[
\cos(2x) = \left(\frac{\sqrt{63}}{8}\right)^2 - \left(\frac{1}{8}\right)^2 = \frac{63}{64} - \frac{1}{64}
\]
\[
\cos(2x) = \frac{62}{64} = \frac{31}{32}
\]
4. **Find \(\tan(2x)\):**
\[
\tan(2x) = \frac{\sin(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b690657-cb09-4dc4-9e96-7e161176d534%2Fd09b01c2-3414-4051-82a5-0d6d3cacd522%2F4przp6b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 3:**
Given that \(\sin x = \frac{1}{8}\) and \(x\) is in quadrant I, use the double-angle identities to find \(\sin(2x)\), \(\cos(2x)\), and \(\tan(2x)\).
*(Hint: find \(\cos x\) and \(\tan x\) first.)*
**Solution Approach:**
1. **Use Pythagorean Identity:**
\[
\sin^2 x + \cos^2 x = 1
\]
Substitute \(\sin x = \frac{1}{8}\):
\[
\left(\frac{1}{8}\right)^2 + \cos^2 x = 1 \implies \frac{1}{64} + \cos^2 x = 1
\]
\[
\cos^2 x = 1 - \frac{1}{64} = \frac{63}{64}
\]
Since \(x\) is in quadrant I:
\[
\cos x = \frac{\sqrt{63}}{8}
\]
2. **Find \(\sin(2x)\) using double-angle identity:**
\[
\sin(2x) = 2 \sin x \cos x
\]
\[
\sin(2x) = 2 \left(\frac{1}{8}\right) \left(\frac{\sqrt{63}}{8}\right) = \frac{\sqrt{63}}{32}
\]
3. **Find \(\cos(2x)\) using double-angle identity:**
\[
\cos(2x) = \cos^2 x - \sin^2 x
\]
\[
\cos(2x) = \left(\frac{\sqrt{63}}{8}\right)^2 - \left(\frac{1}{8}\right)^2 = \frac{63}{64} - \frac{1}{64}
\]
\[
\cos(2x) = \frac{62}{64} = \frac{31}{32}
\]
4. **Find \(\tan(2x)\):**
\[
\tan(2x) = \frac{\sin(
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