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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.

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 dont quite understand how the different values class=

Respuesta :

check the picture below.

[tex]\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}\\\\ -------------------------------\\\\[/tex]

[tex]\bf cos\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1+cos(\beta)}{2}}\implies cos\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1+\frac{4}{5}}{2}} \\\\\\ cos\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{\frac{9}{5}}{2}}\implies cos\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{9}{10}}\implies cos\left( \frac{\beta}{2}\right)=\pm\cfrac{\sqrt{9}}{\sqrt{10}} \\\\\\ cos\left( \frac{\beta}{2}\right)=\pm\cfrac{3}{\sqrt{10}}\implies cos\left( \frac{\beta}{2}\right)=\pm\cfrac{3\sqrt{10}}{10}\\\\ [/tex]
 
[tex]\bf -------------------------------\\\\ sin\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1-cos(\beta)}{2}}\implies sin\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1-\frac{4}{5}}{2}} \\\\\\ sin\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{\frac{1}{5}}{2}}\implies sin\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1}{10}}\implies sin\left( \frac{\beta}{2}\right)=\pm\cfrac{\sqrt{1}}{\sqrt{10}}[/tex]

[tex]\bf sin\left( \frac{\beta}{2}\right)=\pm\cfrac{1}{\sqrt{10}}\implies sin\left( \frac{\beta}{2}\right)=\pm\cfrac{\sqrt{10}}{10}\\\\ -------------------------------\\\\ tan\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1-cos(\beta)}{1+cos(\beta)}}\implies tan\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1-\frac{4}{5}}{1+\frac{4}{5}}}[/tex] 
 
[tex]\bf tan\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{\frac{1}{5}}{\frac{9}{5}}}\implies tan\left( \frac{\beta}{2}\right)=\pm\sqrt{\cfrac{1}{9}}\implies tan\left( \frac{\beta}{2}\right)=\pm \cfrac{1}{3}\\\\ -------------------------------\\\\[/tex]

[tex]\bf cos\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{1+cos(\alpha)}{2}}\implies cos\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{1+\frac{3}{5}}{2}} \\\\\\ cos\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{\frac{8}{5}}{2}}\implies cos\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{4}{5}}\implies cos\left( \frac{\alpha}{2}\right)=\pm\cfrac{\sqrt{4}}{\sqrt{5}} \\\\\\ cos\left( \frac{\alpha}{2}\right)=\pm\cfrac{2}{\sqrt{5}}\implies cos\left( \frac{\alpha}{2}\right)=\pm\cfrac{2\sqrt{5}}{5}\\\\ [/tex]

[tex]\bf -------------------------------\\\\ sin\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{1-cos(\alpha)}{2}}\implies sin\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{1-\frac{3}{5}}{2}} \\\\\\ sin\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{\frac{2}{5}}{2}}\implies sin\left( \frac{\alpha}{2}\right)=\pm\sqrt{\cfrac{1}{5}}\implies sin\left( \frac{\alpha}{2}\right)=\pm\cfrac{\sqrt{1}}{\sqrt{5}} [/tex]

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

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