[tex]\bf \textit{Half-Angle Identities}
\\ \quad \\
sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-cos({{ \theta}})}{2}}\qquad
\boxed{cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}}}[/tex]
[tex]\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}
\\ \quad \\
\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}
\end{cases}\\\\
-----------------------------\\\\[/tex]
[tex]\bf so\qquad
\begin{cases}
2\cdot \cfrac{1}{8}\implies \cfrac{1}{4}\qquad thus\implies \cfrac{\frac{1}{4}}{2}\implies \cfrac{1}{8}\\\\
2\cdot \cfrac{1}{16}\implies \cfrac{1}{8}\qquad thus\implies \cfrac{\frac{1}{8}}{2}\implies \cfrac{1}{16}
\end{cases}\\\\
-----------------------------\\\\[/tex]
[tex]\bf cos\left( \cfrac{\pi }{8} \right)\iff cos\left( \cfrac{\frac{\pi }{4}}{2} \right)
\\\\\\
cos\left( \cfrac{\frac{\pi }{4}}{2} \right)=\pm \sqrt{\cfrac{1+cos\left( \frac{\pi }{4} \right)}{2}}\implies \pm \sqrt{\cfrac{1+\frac{\sqrt{2}}{2}}{2}}\\\\
-----------------------------\\\\
cos\left( \cfrac{\pi }{16} \right)\iff cos\left( \cfrac{\frac{\pi }{8}}{2} \right)
\\\\\\
cos\left( \cfrac{\frac{\pi }{8}}{2} \right)=\pm \sqrt{\cfrac{1+cos\left( \frac{\pi }{8} \right)}{2}}[/tex]
and what is [tex]\bf cos\left( \cfrac{\pi }{8} \right) \ ?[/tex] well, you've just got it from the previous exercise :)
