Answer
$\displaystyle \sin 240^{o}= -\frac{\sqrt{3}}{2}$
$\displaystyle \cos 240^{o}= -\frac{1}{2}$
$\tan 240^{o}= \sqrt{3}$
$\displaystyle \csc 240^{o}= -\frac{2\sqrt{3}}{3}$
$\sec 240^{o}= -2$
$\displaystyle \cot 240^{o}= \frac{\sqrt{3}}{3}$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quadrant: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 0^{o}-\theta
\end{array}\right]$
$240^{o}$ is in quadrant III, for which the reference angle is
$\theta^{\prime}=\theta-180^{o}=240^{o}-180^{o}=60^{o}$
Signs:
In quadrant III, tan and cot are positive, all the others are negative
We use the table Function Values of Special Angles with the corresponding signs:
$\displaystyle \sin 240^{o}=-\sin 60^{o}=-\frac{\sqrt{3}}{2}$
$\displaystyle \cos 240^{o}=-\cos 60^{o}=-\frac{1}{2}$
$\tan 240^{o}=\tan 60^{o}=\sqrt{3}$
$\displaystyle \csc 240^{o}=-\csc 60^{o}=-\frac{2\sqrt{3}}{3}$
$\sec 240^{o}=-\sec 60^{o}=-2$
$\displaystyle \cot 240^{o}=\cot 60^{o}=\frac{\sqrt{3}}{3}$
