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

Question

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 $A.$

Answer 1

$$\tan\left(\arctan\frac{1}{7}+\arctan\frac{3}{4}\right)=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\cdot\frac{3}{4}}=\frac{4+21}{28-3}=1$$ and since $0^{\circ}<\arctan\frac{1}{7}+\arctan\frac{3}{4}<90^{\circ}$, we get the answer: $$\arctan\frac{1}{7}+\arctan\frac{3}{4}=45^{\circ}$$

Answer 2

Hint $$\arctan x +\arctan y =\arctan\left(\frac{x+y}{1-xy}\right) $$ See [Additivity of $\arctan(\frac{x+y}{1-xy})$]

Answer 3

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.