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We are given that \(\cos A = -\frac{15}{17}\) and that \(180^\circ < A < 270^\circ\), meaning that angle \(A\) is in the third quadrant. The goal is to find \(\sin \frac{A}{2}\).

### Steps to solve:

1. **Use the Half-Angle Identity for Sine**:
   The half-angle identity for sine is:
   \[
   \sin \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{2}}
   \]
   The sign depends on the quadrant of \(\frac{A}{2}\). Since \(A\) is in the third quadrant (between \(180^\circ\) and \(270^\circ\)), \(\frac{A}{2}\) is in the second quadrant (between \(90^\circ\) and \(135^\circ\)), where sine is positive. So we will take the positive square root.

2. **Substitute \(\cos A\)**:
   \[
   \sin \frac{A}{2} = \sqrt{\frac{1 - (-\frac{15}{17})}{2}} = \sqrt{\frac{1 + \frac{15}{17}}{2}}
   \]

3. **Simplify the expression**:
   First, simplify the fraction inside the square root:
   \[
   \sin \frac{A}{2} = \sqrt{\frac{\frac{17}{17} + \frac{15}{17}}{2}} = \sqrt{\frac{\frac{32}{17}}{2}} = \sqrt{\frac{32}{34}} = \sqrt{\frac{16}{17}}
   \]

4. **Final result**:
   \[
   \sin \frac{A}{2} = \frac{4}{\sqrt{17}}
   \]
   Rationalizing the denominator:
   \[
   \sin \frac{A}{2} = \frac{4\sqrt{17}}{17}
   \]

Thus, the value of \(\sin \frac{A}{2}\) is \(\frac{4\sqrt{17}}{17}\).

Would you like more details or have any questions?

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Here are 5 related questions for practice:

1. Given \(\cos A = -\frac{15}{17}\), find \(\sin A\).
2. Find \(\tan A\) if \(\cos A = -\frac{15}{17}\) and \(180^\circ < A < 270^\circ\).
3. Solve for \(\sin \frac{A}{2}\) if \(\cos A = \frac{3}{5}\) and \(0^\circ < A < 90^\circ\).
4. What is the exact value of \(\cos \frac{A}{2}\) given \(\cos A = -\frac{15}{17}\) and \(180^\circ < A < 270^\circ\)?
5. Derive the half-angle identity for cosine and apply it to this problem.

**Tip:** Always consider the quadrant when applying trigonometric identities, as it influences the sign of the result.

Given: cos A = -15/17, 180° < A < 270°, Find: sin A/2

Learn how to find sin(A/2) given cos(A) = -15/17 and A between 180° and 270°. This trigonometry problem uses the half-angle identity formula. Suitable for students in Grades 10-12.

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