without using a calculator, compute the sine, cosine, and tangent of 5π/6 by using the reference angle…

Explanation:

Step1: Determine the quadrant

The angle $\frac{5\pi}{6}$ is between $\frac{\pi}{2}=\frac{3\pi}{6}$ and $\pi = \frac{6\pi}{6}$. So it is in the second - quadrant.

Step2: Find the reference angle

The reference angle $\theta_{r}$ for an angle $\theta$ in the second - quadrant is $\theta_{r}=\pi-\theta$. For $\theta = \frac{5\pi}{6}$, $\theta_{r}=\pi-\frac{5\pi}{6}=\frac{\pi}{6}$.

Step3: Find the sine value

In the second - quadrant, $\sin\theta> 0$. Since $\sin\frac{\pi}{6}=\frac{1}{2}$, then $\sin\frac{5\pi}{6}=\frac{1}{2}$.

Step4: Find the cosine value

In the second - quadrant, $\cos\theta<0$. Since $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$, then $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$.

Step5: Find the tangent value

$\tan\theta=\frac{\sin\theta}{\cos\theta}$. So $\tan\frac{5\pi}{6}=\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$.

Answer:

What is the reference angle? $\frac{\pi}{6}$ radians. What quadrant is this angle in? 2 $\sin(\frac{5\pi}{6})=\frac{1}{2}$ $\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}$ $\tan(\frac{5\pi}{6})=-\frac{\sqrt{3}}{3}$