Question

sketch θ = -\frac{2π}{3} radians in standard position on the unit - circle. find the lengths of the legs of its reference triangle. these are labeled a and b in the figure below, when an angle is sketched. then use your reference triangle to find the coordinates of point p. use exact values and not decimal approximations. a = b = p=( )

Explanation:

Step1: Analyze the angle position

The angle $\theta =-\frac{2\pi}{3}$ is in the third - quadrant. Its reference angle $\theta_{r}=\frac{2\pi}{3}-\pi=\frac{\pi}{3}$.

Step2: Find the lengths of the legs of the reference triangle

In a 30 - 60 - 90 triangle (since reference angle is $\frac{\pi}{3}=60^{\circ}$), for a unit - circle (radius $r = 1$), if the hypotenuse of the reference triangle is 1. Let the adjacent side to the reference angle be $a$ and the opposite side be $b$. Using trigonometric relations, $\cos\frac{\pi}{3}=\frac{|a|}{1}$ and $\sin\frac{\pi}{3}=\frac{|b|}{1}$. Since the angle $-\frac{2\pi}{3}$ is in the third - quadrant, $a=-\frac{1}{2}$ and $b =-\frac{\sqrt{3}}{2}$.

Step3: Find the coordinates of point $P$

The coordinates of a point $P(x,y)$ on the unit - circle corresponding to an angle $\theta$ are given by $x = \cos\theta$ and $y=\sin\theta$. For $\theta=-\frac{2\pi}{3}$, $x=\cos(-\frac{2\pi}{3})=-\frac{1}{2}$ and $y = \sin(-\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}$. So $P=(-\frac{1}{2},-\frac{\sqrt{3}}{2})$.

Answer:

$a =-\frac{1}{2}$
$b=-\frac{\sqrt{3}}{2}$
$P=(-\frac{1}{2},-\frac{\sqrt{3}}{2})$