Question 6.3.10: Finding Exact Circular Function Values in Three Ways Find si...
Finding Exact Circular Function Values in Three Ways
Find \sin \frac{5 \pi}{6}, \cos \frac{5 \pi}{6}, \text { and } \tan \frac{5 \pi}{6} by
a. using Figure 47 to find the endpoint of the arc intercepted by t=\frac{5 \pi}{6}.
b. using reference angles and Table 2 from Section 3.
c. converting \frac{5 \pi}{6} to degrees.
TABLE 2 Trigonometric Function Values of Quadrantal Angles
\begin{array}{lccccccc}\hline \theta \text { (degrees) } & \theta \text { (radians) } & \sin \theta & \cos \theta & \tan \theta & \cot \theta & \sec \theta & \csc \theta \\\hline 0^{\circ} & 0 & 0 & 1 & 0 & \text { undefined } & 1 & \text { undefined } \\90^{\circ} & \frac{\pi}{2} & 1 & 0 & \text { undefined } & 0 & \text { undefined } & 1 \\180^{\circ} & \pi & 0 & -1 & 0 & \text { undefined } & -1 & \text { undefined } \\270^{\circ} & \frac{3 \pi}{2} & -1 & 0 & \text { undefined } & 0 & \text { undefined } & -1 \\360^{\circ} & 2 \pi & 0 & 1 & 0 & \text { undefined } & 1 & \text { undefined } \\\hline\end{array}
a. From Figure 47, we see that the endpoint of the arc of length t=\frac{5 \pi}{6} \text { is }\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).
So \cos \frac{5 \pi}{6}=-\frac{\sqrt{3}}{2}, \sin \frac{5 \pi}{6}=\frac{1}{2}, \text { and } \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} \text {. }
b. Either by comparing \frac{5 \pi}{6} \text { to } \frac{6 \pi}{6} or by referring to Figure 47, we see that t=\theta=\frac{5 \pi}{6} radians is in Q II. So its cosine is negative and its sine is positive.The reference angle is
\pi-\frac{5 \pi}{6}=\frac{6 \pi}{6}-\frac{5 \pi}{6}=\frac{\pi}{6}
Then \cos \frac{5 \pi}{6}=-\cos \frac{\pi}{6}=-\frac{\sqrt{3}}{2}, \sin \frac{5 \pi}{6}=\sin \frac{\pi}{6}=\frac{1}{2}, and
\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} .
c. Converting \frac{5 \pi}{6} to degrees, we have \frac{5 \pi}{6} \cdot \frac{180}{\pi}=150^{\circ} . Notice that Figure 47 also gives this information. Because 150° is in Q II, its reference angle is 180^{\circ}-150^{\circ}=30^{\circ} .
Then \cos \frac{5 \pi}{6}=\cos 150^{\circ}=-\cos 30^{\circ}=-\frac{\sqrt{3}}{2},
\sin \frac{5 \pi}{6}=\sin 150^{\circ}=\sin 30^{\circ}=\frac{1}{2}, and
\tan \frac{5 \pi}{6}=\tan 150^{\circ}=\frac{\sin 150^{\circ}}{\cos 150^{\circ}}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}.