Answer:
The answer is D.
Step-by-step explanation:
(1/2, -sq3/2)
Use reference point for pi/3, 11pi/3 is in the fourth quadrant.
By trigonometry functions the coordinates of the terminal point are [tex]$\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$[/tex].
Given:
t = 11 pi / 3
By the definitions of trigonometry functions, the point P has as coordinates P(x, y)
[tex]$x=\cos \left(\frac{11 \pi}{3}\right)$[/tex], and [tex]$y=\sin \left(\frac{11 \pi}{3}\right)$[/tex]
Trig table and unit circle give
[tex]${data-answer}amp;x=\cos \left(\frac{11 \pi}{3}\right)[/tex]
Simplifying the above equation,
[tex]$=\cos \left(-\frac{\pi}{3}+4 \pi\right)[/tex]
[tex]$=\cos \left(-\frac{\pi}{3}\right)[/tex]
[tex]$=\cos \left(\frac{\pi}{3}\right)[/tex]
[tex]$=\frac{1}{2} \\[/tex]
Therefore,
[tex]${data-answer}amp;y=\sin \left(\frac{11 \pi}{3}\right)=\sin \left(-\frac{\pi}{3}+4 \pi\right)[/tex]
Simplifying the above equation,
[tex]$=\sin \left(-\frac{\pi}{3}\right) \\[/tex]
Therefore,
[tex]${data-answer}amp;-\sin \left(\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}[/tex]
[tex]$P\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$[/tex].
P is in Quadrant 4.
[tex]$\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$[/tex] be the coordinates of the terminal point at t = 11 pi / 3.
By trigonometry functions the coordinates of the terminal point are [tex]$\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$[/tex].
Therefore, the correct answer is option D. [tex]$\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$[/tex].
To learn more about trigonometry functions
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