1. **State the problem:** Given that $\sin A = \frac{5}{3}$ and $A$ is in the second quadrant, find $\cos 2A$ and $\sin 2A$. 2. **Note:** The value $\sin A = \frac{5}{3}$ is not possible since sine values must be between -1 and 1. Assuming a typo and the correct value is $\sin A = \frac{3}{5}$. 3. **Recall quadrant rules:** In the second quadrant, sine is positive and cosine is negative. 4. **Find $\cos A$ using Pythagorean identity:** $$\sin^2 A + \cos^2 A = 1$$ $$\left(\frac{3}{5}\right)^2 + \cos^2 A = 1$$ $$\frac{9}{25} + \cos^2 A = 1$$ $$\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25}$$ Since $A$ is in the second quadrant, $\cos A = -\frac{4}{5}$. 5. **Use double angle formulas:** $$\cos 2A = \cos^2 A - \sin^2 A$$ $$\sin 2A = 2 \sin A \cos A$$ 6. **Calculate $\cos 2A$:** $$\cos 2A = \left(-\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$ 7. **Calculate $\sin 2A$:** $$\sin 2A = 2 \times \frac{3}{5} \times \left(-\frac{4}{5}\right) = -\frac{24}{25}$$ **Final answers:** $$\cos 2A = \frac{7}{25}$$ $$\sin 2A = -\frac{24}{25}$$