Question

Last question for today.

I need answers for this, but I also need to know how to solve it. I know the formula, I just don't quite understand how the different values fit into the formula.

Answer 1

check the picture below.


`\bf \textit{Half-Angle Identities} \\ \quad \\ sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-cos({{ \theta}})}{2}}\qquad cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}} \\ \\ \\ tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \boxed{\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 cos\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1+cos(\beta)}{2}}\implies cos\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1+(4)/(5)}{2}} \\\\\\ cos\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{(9)/(5)}{2}}\implies cos\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{9}{10}}\implies cos\left( (\beta)/(2)\right)=\pm\cfrac{√(9)}{√(10)} \\\\\\ cos\left( (\beta)/(2)\right)=\pm\cfrac{3}{√(10)}\implies cos\left( (\beta)/(2)\right)=\pm\cfrac{3√(10)}{10}\\\\`


`\bf -------------------------------\\\\ sin\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1-cos(\beta)}{2}}\implies sin\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1-(4)/(5)}{2}} \\\\\\ sin\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{(1)/(5)}{2}}\implies sin\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1}{10}}\implies sin\left( (\beta)/(2)\right)=\pm\cfrac{√(1)}{√(10)}`


`\bf sin\left( (\beta)/(2)\right)=\pm\cfrac{1}{√(10)}\implies sin\left( (\beta)/(2)\right)=\pm\cfrac{√(10)}{10}\\\\ -------------------------------\\\\ tan\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1-cos(\beta)}{1+cos(\beta)}}\implies tan\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1-(4)/(5)}{1+(4)/(5)}}`


`\bf tan\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{(1)/(5)}{(9)/(5)}}\implies tan\left( (\beta)/(2)\right)=\pm\sqrt{\cfrac{1}{9}}\implies tan\left( (\beta)/(2)\right)=\pm \cfrac{1}{3}\\\\ -------------------------------\\\\`


`\bf cos\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1+cos(\alpha)}{2}}\implies cos\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1+(3)/(5)}{2}} \\\\\\ cos\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{(8)/(5)}{2}}\implies cos\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{4}{5}}\implies cos\left( (\alpha)/(2)\right)=\pm\cfrac{√(4)}{√(5)} \\\\\\ cos\left( (\alpha)/(2)\right)=\pm\cfrac{2}{√(5)}\implies cos\left( (\alpha)/(2)\right)=\pm\cfrac{2√(5)}{5}\\\\`


`\bf -------------------------------\\\\ sin\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1-cos(\alpha)}{2}}\implies sin\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1-(3)/(5)}{2}} \\\\\\ sin\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{(2)/(5)}{2}}\implies sin\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1}{5}}\implies sin\left( (\alpha)/(2)\right)=\pm\cfrac{√(1)}{√(5)}`


`\bf sin\left( (\alpha)/(2)\right)=\pm\cfrac{1}{√(5)}\implies sin\left( (\alpha)/(2)\right)=\pm\cfrac{√(5)}{5}\\\\ -------------------------------\\\\ tan\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1-cos(\alpha)}{1+cos(\alpha)}}\implies tan\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1-(3)/(5)}{1+(3)/(5)}} \\\\\\`


`\bf tan\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{(2)/(5)}{(8)/(5)}}\implies tan\left( (\alpha)/(2)\right)=\pm\sqrt{\cfrac{1}{4}}\implies tan\left( (\alpha)/(2)\right)=\pm \cfrac{1}{2}`