Question

How do you find the point $(x,y)$ on the unit circle that corresponds to the real number $t = \dfrac{{4\pi }}{3}$ ?

Answer 1

Hint:Rectangular coordinate system and polar coordinate system are the two types of coordinate system for plotting a point on the graph paper. The rectangular coordinate system is of the form $(x,y)$ where the distance of this point from the y-axis is denoted by “x” and the distance of the point from the x-axis is denoted by “y” and it is the most commonly used coordinate system.
The polar coordinate system is of the form $(\cos \theta ,\sin \theta )$ where $\theta $ is the counter- clockwise angle between the line joining the point and the origin, and the x-axis. In the question, we are given the value of $\theta $ that is equal to $\dfrac{{4\pi }}{3}$ , so on putting this value in the polar coordinate system, we get the correct answer.

Complete step by step answer:
The coordinates of a circle can also be given as $(\cos \theta ,\sin \theta )$ , in this question we are given $\theta = \dfrac{{4\pi }}{3}$
So,
$
\cos \theta = \cos \dfrac{{4\pi }}{3} = \cos (\pi + \dfrac{\pi }{3}) = - \cos \dfrac{\pi }{3} \\
\Rightarrow \cos \theta = - \dfrac{1}{2} \\
$
And
$
\sin \theta = \sin (\dfrac{{4\pi }}{3}) = \sin (\pi + \dfrac{\pi }{3}) = - \sin \dfrac{\pi }{3} \\
\Rightarrow \sin \theta = - \dfrac{{\sqrt 3 }}{2} \\
$
Hence the point $(x,y)$ on the unit circle that corresponds to the real number $t = \dfrac{{4\pi }}{3}$
is given as $( - \dfrac{1}{2}, - \dfrac{{\sqrt 3 }}{2})$ .

Note: Trigonometric functions are periodic, that is their value repeats after a specific interval of input values but their signs are different in different quadrants. In this question, we are given an angle in the third quadrant. The sign of the sine function and cosine function is negative in the third quadrant that’s why the answer comes out to be negative. For solving any question involving trigonometric ratios, we must know the value of trigonometric ratios at some basic angles.