we have that angle t=[tex] \frac{10 \pi }{3} [/tex]
we know that [tex]\frac{10 \pi }{3}=(\frac{9 \pi }{3}+ \frac{ \pi }{3} )[/tex]
the terminal point belong to the III quadrant so the x-coordinate is negative the y-coordinate is negative
Find the x-coordinate in the unit circle [tex]x=r*cos t[/tex] [tex]r=1 \\ t= \frac{ \pi}{3} \\ x=-1*cos \frac{ \pi }{3} \\ x=- \frac{1}{2} [/tex]
Find the y-coordinate in the unit circle [tex]y=r*sin t[/tex] [tex]r=1 \\ t= \frac{ \pi}{3} \\ y=-1*sin \frac{ \pi }{3} \\ y=- \frac{ \sqrt{3}}{2} [/tex]
the terminal point is [tex](- \frac{1}{2},- \frac{ \sqrt{3}}{2} )[/tex]
therefore
the answer is [tex](- \frac{1}{2},- \frac{ \sqrt{3}}{2} )[/tex]
