Solving the ArcTan of an angle (Radians) by hand?
How do you solve $\arctan(n)$ to radians by hand? I. e. $\arctan(1)$ >> process >> $\pi/4$
::EDIT::
I have this taylor expansion that allows me to calculate an approximate value for arctan, but am wondering if there's a closed-form solution (Or a more general formula than below):
2 Answers
$\begingroup$With a good understanding, you ought to be able to figure out $\mathrm{arctan}(1)$ in your head. Here's what you need to understand:
The tangent function converts angles to slopes. For example, $\mathrm{tan}(37^\circ)$ gives the slope of a line that makes an angle of $37^\circ$ with the $x$-axis.
Therefore, the inverse function (arctangent) converts slopes to angles. For example, $\mathrm{arctan}(2)$ would give the angle between the line $y=2x$ and the $x$-axis.
A line with slope $1$ is inclined at a $45^\circ$ angle. Therefore, $\mathrm{arctan}(1) = 45^\circ$. Converting to radians gives $\mathrm{arctan}(1)=\pi/4$.
Finding the exact arctangent of other values would be much more complicated, though you ought to be able to estimate the arctangent by picturing it. For example, it's easy to estimate that $\mathrm{arctan}(1/3)$ should be about $15$ or $20$ degrees, just by picturing a line with slope $1/3$.
Edit: By the way, if you really want to compute arctangents by hand, one possible method is to use the identity $$ \arctan(x) \;=\; 2\arctan\biggl(\frac{x}{1+\sqrt{1+x^2}}\biggr), $$ which follows from the double-angle formula for tangent. The quantity in parentheses on the right is less than $x/2$, so you can iterate this identity to find a sequence of smaller and smaller angles whose arctangents you want to figure out. (Note that you need to be able to compute square roots by hand.) Once your angle gets small enough, the approximation $$ \arctan(x) \;\approx\; x $$ becomes very accurate.
$\endgroup$ 0 $\begingroup$$\arccos\left(\frac1{\sqrt{1+x^2}}\right)$ is an alternate and an easier function to deal with when using $\arctan(x)$, since you can remember the Taylor series for cosine quite easily.
Take $1-\frac{x^2}2 + \frac{x^4}8 +\ldots$ and plug in $\frac1{\sqrt{1+x^2}}$ for $x$, then set this equal to whatever you want to take the $\arctan$ of, and solve for $x$.
P.S.: this is very labor intensive
Edit: changed wording
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