Find the values of `sin 2 theta, cos 2 theta, and tan 2 theta,` given :
(i) ` sin theta = (3)/(5) , theta ` in Quadrant I.
(ii) `sin theta = 3/5 , theta ` in Quadrant II.
(iii) ` sin theta = - (1)/(2), theta ` in Quadrant IV.
(iv) `tan theta =- (1)/(5) , theta `in Quadrant II.

Answer

To solve the problem of finding the values of \( \sin 2\theta \), \( \cos 2\theta \), and \( \tan 2\theta \) for the given cases, we will follow a systematic approach for each part.

### Part (i): Given \( \sin \theta = \frac{3}{5} \), \( \theta \) in Quadrant I

1. **Find \( \cos \theta \)**:
- Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5} \quad (\text{positive in Quadrant I}) \]

2. **Find \( \tan \theta \)**:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

3. **Calculate \( \sin 2\theta \)**:
\[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \]

4. **Calculate \( \cos 2\theta \)**:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \]

5. **Calculate \( \tan 2\theta \)**:
\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7} \]

### Summary for Part (i):
- \( \sin 2\theta = \frac{24}{25} \)
- \( \cos 2\theta = \frac{7}{25} \)
- \( \tan 2\theta = \frac{24}{7} \)

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### Part (ii): Given \( \sin \theta = \frac{3}{5} \), \( \theta \) in Quadrant II

1. **Find \( \cos \theta \)**:
- \( \cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \)

2. **Find \( \tan \theta \)**:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \]

3. **Calculate \( \sin 2\theta \)**:
\[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{3}{5} \cdot -\frac{4}{5} = -\frac{24}{25} \]

4. **Calculate \( \cos 2\theta \)**:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(-\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \]

5. **Calculate \( \tan 2\theta \)**:
\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{24}{25}}{\frac{7}{25}} = -\frac{24}{7} \]

### Summary for Part (ii):
- \( \sin 2\theta = -\frac{24}{25} \)
- \( \cos 2\theta = \frac{7}{25} \)
- \( \tan 2\theta = -\frac{24}{7} \)

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### Part (iii): Given \( \sin \theta = -\frac{1}{2} \), \( \theta \) in Quadrant IV

1. **Find \( \cos \theta \)**:
- \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \)

2. **Calculate \( \sin 2\theta \)**:
\[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \]

3. **Calculate \( \cos 2\theta \)**:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

4. **Calculate \( \tan 2\theta \)**:
\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3} \]

### Summary for Part (iii):
- \( \sin 2\theta = -\frac{\sqrt{3}}{2} \)
- \( \cos 2\theta = \frac{1}{2} \)
- \( \tan 2\theta = -\sqrt{3} \)

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### Part (iv): Given \( \tan \theta = -\frac{1}{5} \), \( \theta \) in Quadrant II

1. **Find \( \sin \theta \) and \( \cos \theta \)**:
- Let \( \tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{1}{5} \).
- Using the identity \( 1 + \tan^2 \theta = \frac{1}{\cos^2 \theta} \):
\[ 1 + \left(-\frac{1}{5}\right)^2 = \frac{1}{\cos^2 \theta} \implies 1 + \frac{1}{25} = \frac{1}{\cos^2 \theta} \implies \frac{26}{25} = \frac{1}{\cos^2 \theta} \implies \cos^2 \theta = \frac{25}{26} \]
\[ \cos \theta = -\sqrt{\frac{25}{26}} = -\frac{5}{\sqrt{26}} \quad (\text{negative in Quadrant II}) \]
\[ \sin \theta = \tan \theta \cdot \cos \theta = -\frac{1}{5} \cdot -\frac{5}{\sqrt{26}} = \frac{1}{\sqrt{26}} \]

2. **Calculate \( \sin 2\theta \)**:
\[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{1}{\sqrt{26}} \cdot -\frac{5}{\sqrt{26}} = -\frac{10}{26} = -\frac{5}{13} \]

3. **Calculate \( \cos 2\theta \)**:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(-\frac{5}{\sqrt{26}}\right)^2 - \left(\frac{1}{\sqrt{26}}\right)^2 = \frac{25}{26} - \frac{1}{26} = \frac{24}{26} = \frac{12}{13} \]

4. **Calculate \( \tan 2\theta \)**:
\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{5}{13}}{\frac{12}{13}} = -\frac{5}{12} \]

### Summary for Part (iv):
- \( \sin 2\theta = -\frac{5}{13} \)
- \( \cos 2\theta = \frac{12}{13} \)
- \( \tan 2\theta = -\frac{5}{12} \)

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