If cos x = sin 2x cos 2x tan 2x 4 csc x < 0, then ;
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
![This image presents a trigonometric problem, asking to determine the values of the double angle identities based on given conditions.
Given:
- \(\cos x = \frac{4}{5}\)
- \(\csc x < 0\)
Determine:
- \(\sin 2x =\) [Box] ;
- \(\cos 2x =\) [Box] ;
- \(\tan 2x =\) [Box] .
**Explanation:**
To solve this, recognize that \(\csc x < 0\) indicates that \(\sin x\) is negative, implying \(x\) is in the third or fourth quadrant.
1. Use the Pythagorean identity:
\[
\sin^2 x + \cos^2 x = 1
\]
\[
\sin^2 x = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}
\]
Thus, \(\sin x = -\frac{3}{5}\) (since \(\sin x\) is negative).
2. For \(\sin 2x\), use the double angle identity:
\[
\sin 2x = 2 \sin x \cos x = 2 \left(-\frac{3}{5}\right) \left(\frac{4}{5}\right) = -\frac{24}{25}
\]
3. For \(\cos 2x\), use the double angle identity:
\[
\cos 2x = \cos^2 x - \sin^2 x = \left(\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}
\]
4. For \(\tan 2x\), use the identity:
\[
\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{-\frac{24}{25}}{\frac{7}{25}} = -\frac{24}{7}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1da21a78-616d-4fc2-af3c-7f9c768bf6e4%2Facbd913e-d9e1-4626-88c1-06070cd44906%2Fjpzlw2a_processed.png&w=3840&q=75)
Transcribed Image Text:This image presents a trigonometric problem, asking to determine the values of the double angle identities based on given conditions.
Given:
- \(\cos x = \frac{4}{5}\)
- \(\csc x < 0\)
Determine:
- \(\sin 2x =\) [Box] ;
- \(\cos 2x =\) [Box] ;
- \(\tan 2x =\) [Box] .
**Explanation:**
To solve this, recognize that \(\csc x < 0\) indicates that \(\sin x\) is negative, implying \(x\) is in the third or fourth quadrant.
1. Use the Pythagorean identity:
\[
\sin^2 x + \cos^2 x = 1
\]
\[
\sin^2 x = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}
\]
Thus, \(\sin x = -\frac{3}{5}\) (since \(\sin x\) is negative).
2. For \(\sin 2x\), use the double angle identity:
\[
\sin 2x = 2 \sin x \cos x = 2 \left(-\frac{3}{5}\right) \left(\frac{4}{5}\right) = -\frac{24}{25}
\]
3. For \(\cos 2x\), use the double angle identity:
\[
\cos 2x = \cos^2 x - \sin^2 x = \left(\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}
\]
4. For \(\tan 2x\), use the identity:
\[
\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{-\frac{24}{25}}{\frac{7}{25}} = -\frac{24}{7}
\]
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