<s>[INST] How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using the cosine over cosine in square root I got up to
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
 [/INST] I assume you know (or can figure out) $$\cos(-5\pi/4) = -\frac{\sqrt{2}}2,\ \sin(-5\pi/4) = +\frac{\sqrt{2}}2 $$ Applying the half-angle formulas to that, and noting that $-\pi < -5\pi/8  < -\pi/2$
$$ \eqalign{\cos(-5\pi/8) &= - \sqrt{\frac{1+\cos(-5\pi/4)}{2}} = -\frac{\sqrt{2-\sqrt{2}}}{2}\cr
\sin(-5\pi/8) &= - \sqrt{\frac{1-\cos(-5\pi/4)}{2}} = - \frac{\sqrt{2+\sqrt{2}}}{2}}$$
Similarly, since $-\pi/2 < -5\pi/16 < 0$, 
$$ \eqalign{\cos(-5\pi/16) &= +\sqrt{\frac{1+\cos(-5\pi/8)}{2}} = \frac{\sqrt{2-\sqrt{2-\sqrt{2}}}}{2}\cr
\sin(-5\pi/16) &= - \sqrt{\frac{1-\cos(-5\pi/8)}{2}} = - \frac{\sqrt{2+\sqrt{2-\sqrt{2}}}}{2}}$$
so that
$$ \tan(-5\pi/16) = \frac{\sin(-5\pi/16)}{\cos(-5\pi/16)} 
= - \frac{\sqrt{2+\sqrt{2-\sqrt{2}}}}{\sqrt{2-\sqrt{2-\sqrt{2}}}}$$
It turns out (but this is somewhat harder) that you can do some simplification here: you can write it as
$$ \tan(-5\pi/16) = 1 - \sqrt{2} - \sqrt{4-2\sqrt{2}} $$
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