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
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
 A: 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
$$ \tan(-5\pi/16) = 1 - \sqrt{2} - \sqrt{4-2\sqrt{2}} $$
A: Fill in details:
$$r:=\tan\left(-\frac{5\pi}{16}\right)=\tan\frac{11\pi}{16}\implies$$
$$s:=\tan\frac{11\pi}8=\tan\left(2\cdot\frac{11\pi}{16}\right)=\frac{2\tan\frac{11\pi}{16}}{1-\tan^2\frac{11\pi}{16}}=\frac{2r}{1-r^2}$$
and
$$-1=\tan\frac{11\pi}4=\tan\left(2\cdot\frac{11\pi}8\right)=\frac{2s}{1-s^2}\implies$$
$$s^2-2s-1=0\implies s_{1,2}=1\pm\sqrt2$$
and now
$$sr^2+2r-s=0\implies r_{1,2}=\frac{-2\pm\sqrt{4+4s^2}}{2s}=\frac{-1\pm\sqrt{1+s^2}}{s}$$
A: $$\tan{\frac{A}{2}} \equiv \frac{\sin{A}}{1+\cos{A}}$$
$$\tan{\left(-\frac{5\pi}{16}\right)} \equiv \frac{-\sin{\frac{5\pi}{8}}}{1+\cos{\frac{5\pi}{8}}}$$
Solving for the larger angle ratios should be relatively straightforward, however, it will not be as direct due to square roots.
A: Apart from a couple of threes that might be fours ? You are fine upto here
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/4)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/4)}{2}}}}. 
\end{eqnarray*}
Now $\cos(5\pi/4)=\frac{ 1}{\sqrt{2}}$ & after a little bit algebra ...
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 8^{1/4}+\sqrt{\sqrt{2}-1}}{ 8^{1/4}-\sqrt{\sqrt{2}-1}}}. 
\end{eqnarray*}
My Casio fx-83MS gives both sides equal to $-1.4966 \cdots $.
A: The problem is straightforward, it's just a form which is more trouble to rationalize that it's worth. And there are several equivalent expressions depending upon which identities are used.
$$ \tan\left(-\frac{5\pi}{16}\right)=-\frac{\sin\left(\frac{5\pi}{8}\right)}{1+\cos\left(\frac{5\pi}{8}\right)}$$
and 
$$ \sin\left(\frac{5\pi}{8}\right)=\sqrt{\frac{1-\cos(5\pi/4)}{2}}=\frac{1}{2}\sqrt{2+\sqrt{2}}$$
whereas
$$ \cos\left(\frac{5\pi}{8}\right)=-\sqrt{\frac{1+\cos(5\pi/4)}{2}}=-\frac{1}{2}\sqrt{2-\sqrt{2}}$$
Therefore
$$\tan\left(-\frac{5\pi}{16}\right)=\frac{\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}-2} $$
It takes many steps to rationalize this denominator so one may as well leave it in this form.
