Find sin and tan 0 if cos 0=13, assuming that 0 ≤ 0 < π/2. 0 85 sin 0 = tan =

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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**Problem Statement:**

Find \(\sin \theta\) and \(\tan \theta\) if \(\cos \theta = \frac{13}{85}\), assuming that \(0 \leq \theta < \frac{\pi}{2}\).

**Solution:**

1. To find \(\sin \theta\), use the identity \(\sin^2 \theta + \cos^2 \theta = 1\).
   
   \[
   \sin^2 \theta = 1 - \cos^2 \theta
   \]

   Substitute \(\cos \theta = \frac{13}{85}\):

   \[
   \sin^2 \theta = 1 - \left(\frac{13}{85}\right)^2
   \]

   Calculate \(\cos^2 \theta\):

   \[
   \cos^2 \theta = \frac{169}{7225}
   \]

   So,

   \[
   \sin^2 \theta = 1 - \frac{169}{7225} = \frac{7056}{7225}
   \]

   Therefore, \(\sin \theta = \frac{\sqrt{7056}}{\sqrt{7225}} = \frac{84}{85}\).

2. To find \(\tan \theta\), use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

   \[
   \tan \theta = \frac{\frac{84}{85}}{\frac{13}{85}} = \frac{84}{13}
   \]

**Conclusion:**

- \(\sin \theta = \frac{84}{85}\)
- \(\tan \theta = \frac{84}{13}\)
Transcribed Image Text:**Problem Statement:** Find \(\sin \theta\) and \(\tan \theta\) if \(\cos \theta = \frac{13}{85}\), assuming that \(0 \leq \theta < \frac{\pi}{2}\). **Solution:** 1. To find \(\sin \theta\), use the identity \(\sin^2 \theta + \cos^2 \theta = 1\). \[ \sin^2 \theta = 1 - \cos^2 \theta \] Substitute \(\cos \theta = \frac{13}{85}\): \[ \sin^2 \theta = 1 - \left(\frac{13}{85}\right)^2 \] Calculate \(\cos^2 \theta\): \[ \cos^2 \theta = \frac{169}{7225} \] So, \[ \sin^2 \theta = 1 - \frac{169}{7225} = \frac{7056}{7225} \] Therefore, \(\sin \theta = \frac{\sqrt{7056}}{\sqrt{7225}} = \frac{84}{85}\). 2. To find \(\tan \theta\), use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). \[ \tan \theta = \frac{\frac{84}{85}}{\frac{13}{85}} = \frac{84}{13} \] **Conclusion:** - \(\sin \theta = \frac{84}{85}\) - \(\tan \theta = \frac{84}{13}\)
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