Answer :
To determine the coordinates of the terminal point for the given angle \( t = \frac{10\pi}{3} \), let's follow these steps:
1. Normalize the Angle:
The angle \( t = \frac{10\pi}{3} \) is greater than \( 2\pi \). To find the equivalent angle within the interval \( [0, 2\pi) \), we need to subtract multiples of \( 2\pi \) from it until the resulting angle falls within this range:
[tex]\[ \frac{10\pi}{3} - 2\pi = \frac{10\pi}{3} - \frac{6\pi}{3} = \frac{4\pi}{3} \][/tex]
So \( \frac{10\pi}{3} \) is equivalent to \( \frac{4\pi}{3} \) in the standard interval \( [0, 2\pi) \).
2. Determine the Coordinates on the Unit Circle:
The normalized angle \( \frac{4\pi}{3} \) is located in the third quadrant of the unit circle. The reference angle for \( \frac{4\pi}{3} \) can be found by
[tex]\[ \pi - \left( \frac{4\pi}{3} - \pi \right) = \pi - \left( \frac{4\pi}{3} - \frac{3\pi}{3} \right) = \pi - \frac{\pi}{3} = \frac{\pi}{3}. \][/tex]
The coordinates for the angle \( \frac{\pi}{3} \) are \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \). Since \( \frac{4\pi}{3} \) is in the third quadrant, both the \( x \)- and \( y \)-coordinates will be negative:
[tex]\[ \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \][/tex]
Therefore, the coordinates of the terminal point determined by \( t = \frac{10\pi}{3} \) are \(\left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\).
So, the correct answer is:
D. [tex]\(\left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\)[/tex].
1. Normalize the Angle:
The angle \( t = \frac{10\pi}{3} \) is greater than \( 2\pi \). To find the equivalent angle within the interval \( [0, 2\pi) \), we need to subtract multiples of \( 2\pi \) from it until the resulting angle falls within this range:
[tex]\[ \frac{10\pi}{3} - 2\pi = \frac{10\pi}{3} - \frac{6\pi}{3} = \frac{4\pi}{3} \][/tex]
So \( \frac{10\pi}{3} \) is equivalent to \( \frac{4\pi}{3} \) in the standard interval \( [0, 2\pi) \).
2. Determine the Coordinates on the Unit Circle:
The normalized angle \( \frac{4\pi}{3} \) is located in the third quadrant of the unit circle. The reference angle for \( \frac{4\pi}{3} \) can be found by
[tex]\[ \pi - \left( \frac{4\pi}{3} - \pi \right) = \pi - \left( \frac{4\pi}{3} - \frac{3\pi}{3} \right) = \pi - \frac{\pi}{3} = \frac{\pi}{3}. \][/tex]
The coordinates for the angle \( \frac{\pi}{3} \) are \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \). Since \( \frac{4\pi}{3} \) is in the third quadrant, both the \( x \)- and \( y \)-coordinates will be negative:
[tex]\[ \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \][/tex]
Therefore, the coordinates of the terminal point determined by \( t = \frac{10\pi}{3} \) are \(\left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\).
So, the correct answer is:
D. [tex]\(\left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\)[/tex].
