Answer
$\sin{(\frac{7\pi}{6})} = -\frac{1}{2}
\\\cos{\frac{7\pi}{6}=-\frac{\sqrt3}{2}}
\\\tan{\frac{7\pi}{6}}=\frac{\sqrt3}{3}
\\\cot{\frac{7\pi}{6}}=\sqrt3
\\\sec{\frac{7\pi}{6}}=-\frac{2\sqrt3}{3}
\\\csc{\frac{7\pi}{6}}=-2$
Work Step by Step
Recall:
(1) An angle and its reference angle either have the same trigonometric function values or differ only in signs.
(2) $\frac{\pi}{6}$ is a special angle whose sine value is known to be $\frac{1}{2}$ and whose cosine value is known to be $\frac{\sqrt3}{2}$.
(3) $\tan{s} = \frac{\sin{s}}{\cos{s}}$
(4) $\cot{s}= \frac{\cos{s}}{\sin{s}}$
(5) $\sec{s} = \frac{1}{\cos{s}}$
(6) $\csc{s} = \frac{1}{\sin{s}}$
Note that the reference angle of $\frac{7\pi}{6}$ is $\frac{\pi}{6}$.
However, since $\frac{7\pi}{6}$ is in Quadrant III, then its sine and cosine values are both negative .
Therefore,
$\sin{(\frac{7\pi}{6})} = -\frac{1}{2}
\\\cos{\frac{7\pi}{6}=-\frac{\sqrt3}{2}}
\\\tan{\frac{7\pi}{6}}=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\frac{\sqrt3}{3}
\\\cot{\frac{7\pi}{6}}=\dfrac{-\frac{\sqrt3}{3}}{-\frac{1}{2}}=\sqrt3
\\\sec{\frac{7\pi}{6}}=\dfrac{1}{-\frac{\sqrt3}{2}}=-\frac{2\sqrt3}{3}
\\\csc{\frac{7\pi}{6}}=\dfrac{1}{-\frac{1}{2}}=-2$
