Question

find the exact values of the six trigonometric functions of the angle. - 1500°
sin(-1500°)=(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cos(-1500°)=(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
tan(-1500°)=(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cot(-1500°)=(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
csc(-1500°)=(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
sec(-1500°)=(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Find a coterminal angle

Add multiples of 360° to - 1500° to get a positive coterminal angle. -1500°+5×360° = - 1500° + 1800°=300°.

Step2: Recall trig - function values for 300°

We know that for an angle $\theta = 300^{\circ}$, which is in the fourth quadrant.
$\sin(300^{\circ})=-\frac{\sqrt{3}}{2}$, $\cos(300^{\circ})=\frac{1}{2}$, $\tan(300^{\circ})=-\sqrt{3}$, $\cot(300^{\circ})=-\frac{\sqrt{3}}{3}$, $\csc(300^{\circ})=-\frac{2\sqrt{3}}{3}$, $\sec(300^{\circ}) = 2$.
Since the trig - functions of an angle $\theta$ and its coterminal angle have the same values.

Answer:

$\sin(-1500^{\circ})=-\frac{\sqrt{3}}{2}$
$\cos(-1500^{\circ})=\frac{1}{2}$
$\tan(-1500^{\circ})=-\sqrt{3}$
$\cot(-1500^{\circ})=-\frac{\sqrt{3}}{3}$
$\csc(-1500^{\circ})=-\frac{2\sqrt{3}}{3}$
$\sec(-1500^{\circ}) = 2$