Answer
$\sin$(-1860) = $\frac{-\sqrt3}{2}$
$\cos$(-1860) = $\frac{1}{2}$
$\tan$(-1860) = -$\sqrt3$
$\cot$(-1860) = $\frac{-\sqrt3}{3}$
$\csc$(-1860) = $\frac{-2\sqrt3}{3}$
$\sec$(-1860) = 2
Work Step by Step
-1860$^{\circ}$
We must first fine the coterminal angle:
-1860$^{\circ}$ + 360$^{\circ}$ = -1500$^{\circ}$
-1500$^{\circ}$ + 360$^{\circ}$ = -1140$^{\circ}$
-1140$^{\circ}$ + 360$^{\circ}$ = -780$^{\circ}$
-780$^{\circ}$ + 360$^{\circ}$ = -420$^{\circ}$
-420$^{\circ}$ + 360$^{\circ}$ = -60$^{\circ}$
-60$^{\circ}$ + 360$^{\circ}$ = 300$^{\circ}$
150 is in Quadrant IV. Therefore all the trigonometric functions are negative with the exception of $\cos$ and $\sec$.
The reference angle is:
$\theta$$^{1}$ = 360$^{\circ}$ - 300$^{\circ}$ = 60$^{\circ}$
$\sin$(60) = $\frac{-\sqrt3}{2}$
$\cos$(60) = $\frac{1}{2}$
$\tan$(60) = $\frac{-\sqrt3}{1}$ = -$\sqrt3$
$\cot$(60) = $\frac{1}{-\sqrt3}$ = $\frac{-\sqrt3}{3}$
$\csc$(60) = $\frac{2}{-\sqrt3}$ = $\frac{-2\sqrt3}{3}$
$\sec$(60) = $\frac{2}{1}$ = 2
