Respuesta :
[tex]\bf sec(\theta)=\cfrac{hypotenuse}{adjacent}\qquad sec(\theta)=\cfrac{5}{4}\cfrac{\leftarrow hypotenuse=c}{adjacent=a}
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\textit{so, let's find the opposite side "b", using the pythagorean theorem}
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c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{25-16}=b
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\pm 9=b[/tex]
well, the pythagorean theorem, gives us the +/- versions, so, which one is it?
well, our angle is at 0° < θ < 90°, meaning the 1st quadrant, and there, the opposite side, or "y" or "b", is positive, so is b = 9 then
now
[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}} \\ \quad \\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{1-cos({{ \theta}})}{sin({{ \theta}})} \end{cases}[/tex]
thus
[tex]\bf sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-sin({{ \theta}})}{2}}\implies sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-\frac{4}{5}}{2}} \\\\\\ sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{\frac{1}{5}}{2}}\implies sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1}{10}}[/tex]
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[tex]\bf cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+\frac{4}{5}}{2}} \\\\\\ cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{\frac{9}{5}}{2}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{9}{10}}[/tex]
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[tex]\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{\frac{9}{5}}{1+\frac{4}{5}}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{\frac{9}{5}}{\frac{9}{5}} \\\\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)=1[/tex]
well, the pythagorean theorem, gives us the +/- versions, so, which one is it?
well, our angle is at 0° < θ < 90°, meaning the 1st quadrant, and there, the opposite side, or "y" or "b", is positive, so is b = 9 then
now
[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}} \\ \quad \\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{1-cos({{ \theta}})}{sin({{ \theta}})} \end{cases}[/tex]
thus
[tex]\bf sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-sin({{ \theta}})}{2}}\implies sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-\frac{4}{5}}{2}} \\\\\\ sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{\frac{1}{5}}{2}}\implies sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1}{10}}[/tex]
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[tex]\bf cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+\frac{4}{5}}{2}} \\\\\\ cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{\frac{9}{5}}{2}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{9}{10}}[/tex]
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[tex]\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{\frac{9}{5}}{1+\frac{4}{5}}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{\frac{9}{5}}{\frac{9}{5}} \\\\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)=1[/tex]
