Answer
$\sin$(-1020) = $\frac{\sqrt3}{2}$
$\cos$(-1020) = $\frac{1}{2}$
$\tan$(-1020) = $\sqrt3$
$\cot$(-1020) = $\frac{2\sqrt3}{3}$
$\csc$(-1020) = 2
$\sec$(-1020) = $\frac{\sqrt3}{3}$
Work Step by Step
-1020$^{\circ}$
We must first fine the coterminal angle:
-1020$^{\circ}$ + 360$^{\circ}$ = -660$^{\circ}$
-660$^{\circ}$ + 360$^{\circ}$ = -300$^{\circ}$
-300$^{\circ}$ + 360$^{\circ}$ = 60$^{\circ}$
-150 is in Quadrant I. Therefore all the trigonometric functions are positive.
The reference angle is:
$\theta$$^{1}$ = 60$^{\circ}$
$\sin$(60) = $\frac{\sqrt3}{2}$
$\cos$(60) = $\frac{1}{2}$
$\tan$(60) = $\sqrt3$
$\cot$(60) = $\frac{2\sqrt3}{3}$
$\csc$(60) = 2
$\sec$(60) = $\frac{\sqrt3}{3}$
