Turn a real number (with a complex closed form) into its trigonometric form This post outlines 'fake' complex numbers (real numbers with complex closed form that usually come from the roots of unfactorable cubics (the example I need right now), or they can come from things like $i^i = e^{-\frac{\pi}{2}}$), in his own answer he gives the example: $\sqrt[3]{1+i \sqrt{7}}+\sqrt[3]{1-i \sqrt{7}}$, which from first glance doesn't look like it could simplified any futher, then states:

Using Euler's formula we can find a trigonometric representation of this number. $2 \sqrt{2} \cos{\left(\frac{\tan^{-1}{\left(\sqrt{7}\right)}}{3}\right)}$

Which when put into WolframAlpha is confirmed, they are both approximately equal to $2.602$
How did he do that? Euler's formula, $e^{ix} = \cos(x) + i\sin(x)$ doesn't seem relevant, the example has no clear $a+ib$ form (and it shouldn't because it is a real number, $b$ will just be $0$, so a will just be the answer anyway) and there is an inverse tangent present in the expression, also covered by Euler's formula somehow? How does he just go from to $A$ to $B$ like it's common knowledge?
Question applies to the example but it would be nice to have further advice on how to turn other fake complex numbers into real forms like this without WolframAlpha.

Recall DeMoivre's formula:
$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
$$\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$$
This can trivially be derived from Euler's formula.
Also note that in polar coordinates,
$$1 + i\sqrt{7} = \sqrt{8} \operatorname{cis}(\tan^{-1}\sqrt{7})$$
So by DeMoivre's formula:
$$(1 + i\sqrt{7})^{1/3}$$
$$ = \left(\sqrt{8} \operatorname{cis}(\tan^{-1}\sqrt{7})\right)^{1/3}$$
$$ = 8^{1/6} \operatorname{cis}(\frac{1}{3}\tan^{-1}\sqrt{7})$$
$$ = \sqrt{2} \operatorname{cis}(\frac{\tan^{-1}\sqrt{7}}{3})$$
$$ = \sqrt{2} \cos(\frac{\tan^{-1}\sqrt{7}}{3}) + i\sqrt{2}\sin(\frac{\tan^{-1}\sqrt{7}}{3}) $$
Similarly,
$$(1 - i\sqrt{7})^{1/3}$$
$$ = \left(\sqrt{8} \operatorname{cis}(-\tan^{-1}\sqrt{7})\right)^{1/3}$$
$$ = 8^{1/6} \operatorname{cis}(-\frac{1}{3}\tan^{-1}\sqrt{7})$$
$$ = \sqrt{2} \operatorname{cis}(-\frac{\tan^{-1}\sqrt{7}}{3})$$
$$ = \sqrt{2} \cos(-\frac{\tan^{-1}\sqrt{7}}{3}) + i\sqrt{2}\sin(-\frac{\tan^{-1}\sqrt{7}}{3}) $$
$$ = \sqrt{2} \cos(\frac{\tan^{-1}\sqrt{7}}{3}) - i\sqrt{2}\sin(\frac{\tan^{-1}\sqrt{7}}{3}) $$
So adding the two cube roots gives $2\sqrt{2} \cos(\frac{\tan^{-1}\sqrt{7}}{3})$, QED.

More generally, if $z = r\operatorname{cis}\theta$, then
$$\sqrt[n]{z} + \sqrt[n]{\overline{z}} = 2\sqrt[n]{r} \cos\left(\frac{\theta}{n}\right)$$
Or if $z = x + iy$, then:
$$\sqrt[n]{x + iy} + \sqrt[n]{x - iy} = 2\sqrt[2n]{x^2 + y^2} \cos\left(\frac{\arg(x + iy)}{n}\right)$$
