Answer
$\sin$(-1290) = $\frac{1}{2}$
$\cos$(-1290) = -$\frac{\sqrt3}{2}$
$\tan$(-1290) = -$\frac{\sqrt3}{3}$
$\cot$(-1290) = 2
$\csc$(-1290) = -$\frac{2\sqrt3}{3}$
$\sec$(-1290) = -$\sqrt3$
Work Step by Step
-1290$^{\circ}$
We must first fine the coterminal angle:
-1290$^{\circ}$ + 360$^{\circ}$ = -930$^{\circ}$
-930$^{\circ}$ + 360$^{\circ}$ = -570$^{\circ}$
-570$^{\circ}$ + 360$^{\circ}$ = -210$^{\circ}$
-210$^{\circ}$ + 360$^{\circ}$ = 150$^{\circ}$
150 is in Quadrant II. Therefore all the trigonometric functions are negative with the exception of $\sin$ and $\csc$.
The reference angle is:
$\theta$$^{1}$ = 180$^{\circ}$ - 150$^{\circ}$ = 30$^{\circ}$
$\sin$(30) = $\frac{1}{2}$
$\cos$(30) = -$\frac{\sqrt3}{2}$
$\tan$(30) = -$\frac{\sqrt3}{3}$
$\cot$(30) = 2
$\csc$(30) = -$\frac{2\sqrt3}{3}$
$\sec$(30) = -$\sqrt3$
