Respuesta :

ujalakhan18

Answer:

P([tex]\frac{5\pi }{6}[/tex]) = ( -[tex]\sqrt{3} / 2[/tex] , [tex]1/2[/tex] )

Step-by-step explanation:

Cos [tex]\frac{5\pi }{6}[/tex] = [tex]\sqrt{3} / 2[/tex]

Sin [tex]\frac{5\pi }{6}[/tex] = [tex]1/2[/tex]

Since [tex]\frac{5\pi }{6}[/tex] lies in Quadrant II , So Cos [tex]\frac{5\pi }{6}[/tex] < 0, Sin [tex]\frac{5\pi }{6}[/tex] > 0

So,

The coordinates are P([tex]\frac{5\pi }{6}[/tex]) = ( -[tex]\sqrt{3} / 2[/tex] , [tex]1/2[/tex] )

09pqr4sT

Answer:

[tex]p(\frac{5 \pi}{6} ) = (- \frac{\sqrt{3} }{2} , \frac{1}{2} )[/tex]

Step-by-step explanation:

[tex]cos(\frac{5 \pi}{6} )=\frac{\sqrt{3} }{2}[/tex]

[tex]sin(\frac{5 \pi}{6} ) = \frac{1}{2}[/tex]

(5π)/6 is in quadrant II, so..

[tex]cos(\frac{5 \pi}{6} ) < 0[/tex]

[tex]sin(\frac{5 \pi}{6} ) > 0[/tex]

The points are:

[tex]p(\frac{5 \pi}{6} ) = (- \frac{\sqrt{3} }{2} , \frac{1}{2} )[/tex]

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