The problem you provided is:

Given: $\tan A = \frac{4}{3}$, with $180^\circ < A < 270^\circ$ (i.e., in the third quadrant).

Find: $\sin \frac{A}{2}$.

Solution:

We will use the following steps:

1. Determine $\sin A$ and $\cos A$:

   • Since $\tan A = \frac{\sin A}{\cos A}$, and $\tan A = \frac{4}{3}$, we can use a right triangle to find $\sin A$ and $\cos A$.
   • In the third quadrant, both sine and cosine are negative.

   Let: $\sin A = -\frac{4}{5}, \quad \cos A = -\frac{3}{5}$ (These values come from using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$).

2. Apply the half-angle formula for sine: The half-angle identity for sine is: $\sin \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{2}}$ Since $180^\circ < A < 270^\circ$, $\frac{A}{2}$ is in the second quadrant, where sine is positive. Thus, we use the positive square root.

3. Substitute $\cos A = -\frac{3}{5}$ into the half-angle formula: $\sin \frac{A}{2} = \sqrt{\frac{1 - (-\frac{3}{5})}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}}$

4. Simplify: $\sin \frac{A}{2} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

Thus, $\sin \frac{A}{2} = \frac{2\sqrt{5}}{5}$.

Would you like further details on any step? Here are some related questions:

1. How do you find $\cos A$ and $\sin A$ using the Pythagorean identity?
2. Why is the half-angle formula used here, and how does it apply to this problem?
3. Can you explain how to determine the correct sign for the half-angle formula?
4. What is the relationship between the quadrant and the sign of $\sin \frac{A}{2}$?
5. How would this solution change if $A$ were in a different quadrant?

Tip: Always consider the quadrant to decide the correct sign of trigonometric values!