<s>[INST] What is the value of $\operatorname{arctan} \left(-\frac{12}{5}\right)$? I am trying to find the value of $\sin(\operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $\operatorname{tan}(\theta) = -12/5$. But this is not one of the easy angles to simply look up the value. 
I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example. 
 [/INST] Let $\theta = \arctan \frac {-12}{5}$
$\tan \theta = \frac {-12}{5} = \frac {\sin \theta}{\cos \theta}$ which means there there is a right triangle with angle $\theta$ and the opposite side = $r\sin \theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r \cos \theta = 5$ where $r$ is the hypotenuse.  
So $\sin \theta = \frac {-12}r$ where $r$ is the hypotenuse = $\sqrt {(-12)^2 + 5^2} = 13$.
So $\sin \theta = \frac {-12}{13}$.
....
In general:
$\sin (\arcsin \frac ab) = \frac {a}{\sqrt{a^2 + b^2}}$
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