based on the unit circle shown, josiah claims that sin(\\(\\frac{5\\pi}{6}\\)) = -\\(\\frac{\\sqrt{3}}{2}\\)…
based on the unit circle shown, josiah claims that sin(\\(\\frac{5\\pi}{6}\\)) = -\\(\\frac{\\sqrt{3}}{2}\\). is josiah correct? use the drop - down menus to explain.
Answer
Answer:
No
Explanation:
Step1: Recall sine - unit circle relationship
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: Determine the quadrant of $\frac{5\pi}{6}$
The angle $\theta=\frac{5\pi}{6}$ is in the second - quadrant ($\frac{\pi}{2}<\frac{5\pi}{6}<\pi$). In the second - quadrant, the $y$ - coordinate of a point on the unit circle is positive.
Step3: Find the value of $\sin(\frac{5\pi}{6})$
We know that for $\theta = \frac{5\pi}{6}$, the reference angle $\theta_{r}=\pi-\frac{5\pi}{6}=\frac{\pi}{6}$. And $\sin(\frac{\pi}{6})=\frac{1}{2}$. Since $\sin\theta> 0$ in the second - quadrant, $\sin(\frac{5\pi}{6})=\frac{1}{2}\neq-\frac{\sqrt{3}}{2}$.