1. **Problem statement:** Given $\sin A = \frac{4}{5}$ and $\cos B = \frac{5}{3}$ (note: $\cos B$ cannot be $\frac{5}{3}$ since cosine values must be between -1 and 1, so we assume a typo and treat it as $\cos B = \frac{3}{5}$). We need to evaluate $\sin (A + B)$ and $\cos (A - B)$ without using a table. 2. **Formulas used:** - $\sin (A + B) = \sin A \cos B + \cos A \sin B$ - $\cos (A - B) = \cos A \cos B + \sin A \sin B$ 3. **Find missing values:** - Since $\sin A = \frac{4}{5}$, use Pythagorean identity $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$. - Since $\cos B = \frac{3}{5}$, use $\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$. 4. **Calculate $\sin (A + B)$:** $$\sin (A + B) = \sin A \cos B + \cos A \sin B = \frac{4}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{4}{5} = \frac{12}{25} + \frac{12}{25} = \frac{24}{25}$$ 5. **Calculate $\cos (A - B)$:** $$\cos (A - B) = \cos A \cos B + \sin A \sin B = \frac{3}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{4}{5} = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$$ **Final answers:** - $\sin (A + B) = \frac{24}{25}$ - $\cos (A - B) = 1$