based on the unit circle shown, josiah claims that sin(5π/6)= -√3/2. is josiah correct? use the drop - down…

Explanation:

Step1: Recall unit - circle definition of sine

On the unit circle, for an angle $\theta$, $\sin\theta$ is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.

Step2: Analyze the angle $\frac{5\pi}{6}$

The angle $\theta=\frac{5\pi}{6}$ is in the second - quadrant. The reference angle for $\frac{5\pi}{6}$ is $\pi-\frac{5\pi}{6}=\frac{\pi}{6}$. In the second - quadrant, the $x$-coordinate is negative and the $y$-coordinate is positive. For an angle with reference angle $\frac{\pi}{6}$, the coordinates of the point on the unit circle are $(-\frac{\sqrt{3}}{2},\frac{1}{2})$. Since $\sin\theta$ is the $y$-coordinate of the point on the unit circle corresponding to the angle $\theta$, for $\theta = \frac{5\pi}{6}$, $\sin(\frac{5\pi}{6})=\frac{1}{2}$.

Answer:

Josiah is incorrect. The $y$-coordinate of the endpoint of the terminal side is $\frac{1}{2}$.