Suppose that sin0 COS tan 2 Find the exact values of cos and tan 2 0 2 = 0 = = - 0 5 13 and 0°<0<90°. 0 2 8 Undefined X √67 S

Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
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### Trigonometric Identities and Half-Angle Formulas

**Problem Statement:**
Suppose that \(\sin \theta = \frac{5}{13}\) and \(0^\circ < \theta < 90^\circ\).

Find the exact values of \(\cos \frac{\theta}{2}\) and \(\tan \frac{\theta}{2}\).

**Solution:**

1. **Given:**
   \(\sin \theta = \frac{5}{13}\)

2. **Objective:**
   Find:
   \[
   \cos \frac{\theta}{2} \quad \text{and} \quad \tan \frac{\theta}{2}
   \]

3. **Using the Half-Angle Formulas:**

   For \(\cos \frac{\theta}{2}\):
   \[
   \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}
   \]
   
   For \(\tan \frac{\theta}{2}\):
   \[
   \tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}
   \]

4. **Find \(\cos \theta\):**
   \[
   \cos^2 \theta + \sin^2 \theta = 1
   \]
   \[
   \cos^2 \theta + \left(\frac{5}{13}\right)^2 = 1
   \]
   \[
   \cos^2 \theta + \frac{25}{169} = 1
   \]
   \[
   \cos^2 \theta = 1 - \frac{25}{169}
   \]
   \[
   \cos^2 \theta = \frac{144}{169}
   \]
   \[
   \cos \theta = \pm \frac{12}{13}
   \]

   Since \(0^\circ < \theta < 90^\circ\), \(\cos \theta\) is positive:
   \[
   \cos \theta = \frac{12}{13}
   \]

5. **Calculate \(\cos \frac{\theta}{2}\):**
   \[
   \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \
Transcribed Image Text:### Trigonometric Identities and Half-Angle Formulas **Problem Statement:** Suppose that \(\sin \theta = \frac{5}{13}\) and \(0^\circ < \theta < 90^\circ\). Find the exact values of \(\cos \frac{\theta}{2}\) and \(\tan \frac{\theta}{2}\). **Solution:** 1. **Given:** \(\sin \theta = \frac{5}{13}\) 2. **Objective:** Find: \[ \cos \frac{\theta}{2} \quad \text{and} \quad \tan \frac{\theta}{2} \] 3. **Using the Half-Angle Formulas:** For \(\cos \frac{\theta}{2}\): \[ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \] For \(\tan \frac{\theta}{2}\): \[ \tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \] 4. **Find \(\cos \theta\):** \[ \cos^2 \theta + \sin^2 \theta = 1 \] \[ \cos^2 \theta + \left(\frac{5}{13}\right)^2 = 1 \] \[ \cos^2 \theta + \frac{25}{169} = 1 \] \[ \cos^2 \theta = 1 - \frac{25}{169} \] \[ \cos^2 \theta = \frac{144}{169} \] \[ \cos \theta = \pm \frac{12}{13} \] Since \(0^\circ < \theta < 90^\circ\), \(\cos \theta\) is positive: \[ \cos \theta = \frac{12}{13} \] 5. **Calculate \(\cos \frac{\theta}{2}\):** \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \
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