4. if the angle $\theta=\frac{5pi}{6}$ is in standard position on the unit circle, which ordered pair does…

4. if the angle $\theta=\frac{5pi}{6}$ is in standard position on the unit circle, which ordered pair does its terminal side pass through?\n$(-\\frac{1}{2},\\frac{\\sqrt{3}}{2})$\n$(-\\frac{\\sqrt{3}}{2},\\frac{1}{2})$\n$(-\\frac{1}{2},-\\frac{\\sqrt{3}}{2})$\n$(-\\frac{\\sqrt{3}}{2},-\\frac{1}{2})$\n

4. if the angle $\theta=\frac{5pi}{6}$ is in standard position on the unit circle, which ordered pair does its terminal side pass through?\n$(-\\frac{1}{2},\\frac{\\sqrt{3}}{2})$\n$(-\\frac{\\sqrt{3}}{2},\\frac{1}{2})$\n$(-\\frac{1}{2},-\\frac{\\sqrt{3}}{2})$\n$(-\\frac{\\sqrt{3}}{2},-\\frac{1}{2})$\n

Answer

Explanation:

Step1: Recall the unit - circle formula

For an angle (\theta) in standard position on the unit circle, the terminal side passes through the point ((\cos\theta,\sin\theta)).

Step2: Calculate (\cos\theta)

Given (\theta=\frac{5\pi}{6}), (\cos\frac{5\pi}{6}=\cos(\pi - \frac{\pi}{6})). Using the formula (\cos(A - B)=\cos A\cos B+\sin A\sin B) (here (A = \pi), (B=\frac{\pi}{6})), (\cos(\pi-\alpha)=-\cos\alpha). So (\cos\frac{5\pi}{6}=-\cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}).

Step3: Calculate (\sin\theta)

(\sin\frac{5\pi}{6}=\sin(\pi-\frac{\pi}{6})). Using the formula (\sin(A - B)=\sin A\cos B-\cos A\sin B) (here (A = \pi), (B=\frac{\pi}{6})), (\sin(\pi-\alpha)=\sin\alpha). So (\sin\frac{5\pi}{6}=\sin\frac{\pi}{6}=\frac{1}{2}).

Answer:

(\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right))