Respuesta :

Recall that the coordinates of the terminal points are

[tex]x=\cos \theta\text{ and y=sin}\theta[/tex][tex]\text{Given that }\theta=\frac{5\pi}{6}\text{ radian.}[/tex][tex]\text{Substitue }\theta=\frac{5\pi}{6}\text{ in x=cos}\theta\text{ to compute x.}[/tex][tex]x=\cos (\frac{5\pi}{6})[/tex]

[tex]=\cos (\pi-\frac{\pi}{6})[/tex][tex]\text{Use cos(}\pi-\theta)=-\cos \theta\text{.}[/tex]

[tex]=-\cos (\frac{\pi}{6})[/tex]

[tex]Use\text{ }\cos (\frac{\pi}{6})=\frac{\sqrt[]{3}}{2}[/tex][tex]x=-\frac{\sqrt[]{3}}{2}[/tex]

[tex]\text{Substitue }\theta=\frac{5\pi}{6}\text{ in y=sin}\theta\text{ to compute y.}[/tex][tex]y=\sin (\frac{5\pi}{6})[/tex][tex]=\sin (\pi-\frac{\pi}{6})[/tex]

[tex]\text{Use sin(}\pi-\theta)=\sin \theta.[/tex]

[tex]=\sin (\frac{\pi}{6})[/tex][tex]Use\text{ }\sin (\frac{\pi}{6})=\frac{1}{2}\text{.}[/tex][tex]y=\frac{1}{2}[/tex]

Hence the terminal point is

[tex](x,y)=(-\frac{\sqrt[]{3}}{2},\frac{1}{2})[/tex]

Option third is correct.

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