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{(1)/(4)}{2}\implies \cfrac{1}{8}\\\\ 2\cdot \cfrac{1}{16}\implies \cfrac{1}{8}\qquad thus\implies \cfrac{(1)/(8)}{2}\implies \cfrac{1}{16} \end{cases}\\\\ -----------------------------\\\\`


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

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