If sin A = 12/13 and Cos B = 3/5, Evaluate a) Sin (A + B) b) Cos (A + B) c) Sin 2B

Here are the evaluations for the trigonometric expressions:

First, we need to find the missing trigonometric values for angles A and B using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$. We assume A and B are acute angles, so their sine and cosine values are positive.

Given: $\sin A = \frac{12}{13}$ and $\cos B = \frac{3}{5}$.

Step 1: Find $\cos A$.

$$\cos^2 A = 1 - \sin^2 A$$ $$\cos^2 A = 1 - \left(\frac{12}{13}\right)^2$$ $$\cos^2 A = 1 - \frac{144}{169}$$ $$\cos^2 A = \frac{169 - 144}{169}$$ $$\cos^2 A = \frac{25}{169}$$ $$\cos A = \sqrt{\frac{25}{169}} = \frac{5}{13}$$

Step 2: Find $\sin B$.

$$\sin^2 B = 1 - \cos^2 B$$ $$\sin^2 B = 1 - \left(\frac{3}{5}\right)^2$$ $$\sin^2 B = 1 - \frac{9}{25}$$ $$\sin^2 B = \frac{25 - 9}{25}$$ $$\sin^2 B = \frac{16}{25}$$ $$\sin B = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

Now we have all the necessary values:
$\sin A = \frac{12}{13}$
$\cos A = \frac{5}{13}$
$\sin B = \frac{4}{5}$
$\cos B = \frac{3}{5}$