Answer:
Terminal point will be (-\(\frac{1}{2}\), -\(\frac{\sqrt{3} }{2}\))
Step-by-step explanation:
Terminal point (x, y) on a circle is defined by P(x, y) = ( r cost, r sin t)
Where t is the angle = \(\frac{4\pi }{3}\)
and Radius of the circle = 1 unit [unit circle]
x coordinate = r cost
= \(1\times cos\frac{4\pi }{3}\)
= \(cos(\frac{3\pi }{2}-\frac{\pi }{3})\)
= -\((1\times cos\frac{\pi }{3})\) [angle lies in third quadrant so the value will be negative]
= -\(\frac{1}{2}\)
y- coordinate = r sint
= \(1\times sin\frac{4\pi }{3}\)
= -\(sin(\frac{2\pi }{3}-\frac{\pi }{3})\)
= -\((1\times sin\frac{\pi }{3})\)
= -\(\frac{\sqrt{3} }{2}\)
Therefore, terminal point will be (-\(\frac{1}{2}\), -\(\frac{\sqrt{3} }{2}\))
