### Instruction: If $e^{i \alpha} = \frac{3}{5}  +\frac{4}{5} i$ and $e^{i \beta} = -\frac{12}{13} + \frac{5}{13} i,$ then find $\sin (\alpha + \beta).$.### Response: Ok, the first complex number is $3/5 + 4/5i$.
The second one is $-12/13 + 5/13i$.
Right, so we need to find the sine of their sum.
Let's first get $\alpha$. We have that $e^{i \alpha} = 3/5 + 4/5i$.
So, $\cos \alpha = 3/5$ and $\sin \alpha = 4/5$.
Now let's get $\beta$. We have that $e^{i \beta} = -12/13 + 5/13 i$.
So, $\cos \beta = -12/13$ and $\sin \beta = 5/13$.
Now we can get $\sin (\alpha + \beta)$. That would be $\sin \alpha \cos \beta + \cos \alpha \sin \beta$.
So we have $(3/5)(5/13) + (4/5)(-12/13) = 15/65 - 48/65 = -33/65.$