Question

Find the exact values of sin(θ/2), cos(θ/2), and tan(θ/2) for the given conditions.

sec θ =
5
4
; 0° < θ < 90°

sin(θ/2) =
cos(θ/2) =
tan(θ/2) =

Answer 1

`\bf sec(\theta)=\cfrac{hypotenuse}{adjacent}\qquad sec(\theta)=\cfrac{5}{4}\cfrac{\leftarrow hypotenuse=c}{adjacent=a} \\\\\\ \textit{so, let's find the opposite side`

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


`\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}`

thus


`\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-(4)/(5)}{2}} \\\\\\ sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{(1)/(5)}{2}}\implies sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1}{10}}`
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`\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+(4)/(5)}{2}} \\\\\\ cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{(9)/(5)}{2}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{9}{10}}`

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`\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{(9)/(5)}{1+(4)/(5)}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{(9)/(5)}{(9)/(5)} \\\\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)=1`