Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$ 
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$

I want to compute this sum by computing one term at a time. It's clear that 
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right triangle with sides $1$, $7$ and $5\sqrt{2},$ I get that $$\begin{array}{lcl}
\sin{A} & = & \frac{1}{5\sqrt{2}}\Longleftrightarrow A= \arcsin{\frac{1}{5\sqrt{2}}} \\
\cos{A} & = & \frac{7}{5\sqrt{2}}\Longleftrightarrow A= \arccos{\frac{7}{5\sqrt{2}}} \\
\end{array}$$
But this will not get me standard angles for $A.$

I think of these sorts of things in terms of complex numbers.  Let $\theta = \arctan 1/7$.  Then $e^{i\theta} = \cos \theta + i \sin \theta$.  We need $\sin \theta / \cos \theta = 1/7$, so this will be a multiple of $7 + i$, but it must have length 1, so $e^{i \theta} = (7+i)/\sqrt{50}$.  Similarly let $\phi = \arctan 3/4$; by the same argument $e^{i \phi} = (4 + 3i)/5$.  Then we have
$$e^{i(\theta + \phi)} = {(7+i) (4+3i) \over 5 \sqrt{50}} = {28 + 4i + 21i - 3  \over 25 \sqrt{2}} = {25 + 25i \over 25 \sqrt{2}} = {1+i \over \sqrt{2}}$$
from which we can see that $\theta + \phi = \arctan 1 = \pi/4$.
This illustrates the general point that lots of trig identities are in some sense just identities about complex numbers.
