Respuesta :

[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 :)

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