Question

Can someone help me with b and c?

Answer 1

$\bf \textit{Half-Angle Identities} \\ \quad \\ sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-cos({{ \theta}})}{2}}\qquad \boxed{cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}}}$

$\bf tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{sin({{ \theta}})}{1+cos({{ \theta}})} \\ \quad \\ \cfrac{1-cos({{ \theta}})}{sin({{ \theta}})} \end{cases}\\\\ -----------------------------$

$\bf so\qquad \begin{cases} 2\cdot \cfrac{1}{8}\implies \cfrac{1}{4}\qquad thus\implies \cfrac{\frac{1}{4}}{2}\implies \cfrac{1}{8}\\\\ 2\cdot \cfrac{1}{16}\implies \cfrac{1}{8}\qquad thus\implies \cfrac{\frac{1}{8}}{2}\implies \cfrac{1}{16} \end{cases}\\\\ -----------------------------$

$\bf cos\left( \cfrac{\pi }{8} \right)\iff cos\left( \cfrac{\frac{\pi }{4}}{2} \right) \\\\\\ cos\left( \cfrac{\frac{\pi }{4}}{2} \right)=\pm \sqrt{\cfrac{1+cos\left( \frac{\pi }{4} \right)}{2}}\implies \pm \sqrt{\cfrac{1+\frac{\sqrt{2}}{2}}{2}}\\\\ -----------------------------\\\\ cos\left( \cfrac{\pi }{16} \right)\iff cos\left( \cfrac{\frac{\pi }{8}}{2} \right) \\\\\\ cos\left( \cfrac{\frac{\pi }{8}}{2} \right)=\pm \sqrt{\cfrac{1+cos\left( \frac{\pi }{8} \right)}{2}}$

and what is $\bf cos\left( \cfrac{\pi }{8} \right) \ ?$   well, you've just got it from the previous exercise :)