Question

find the exact values of the six trigonometric functions of the angle. - 690°
sin(-690°)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cos(-690°)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
tan(-690°)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cot(-690°)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
csc(-690°)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
sec(-690°)= (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 \(360^{\circ}\) twice to \(- 690^{\circ}\). \(-690^{\circ}+2\times360^{\circ}=-690^{\circ} + 720^{\circ}=30^{\circ}\). Trigonometric functions of coterminal angles are equal.

Step2: Calculate \(\sin(-690^{\circ})\)

Since \(\sin(-690^{\circ})=\sin(30^{\circ})\), and \(\sin(30^{\circ})=\frac{1}{2}\).

Step3: Calculate \(\cos(-690^{\circ})\)

Since \(\cos(-690^{\circ})=\cos(30^{\circ})\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\).

Step4: Calculate \(\tan(-690^{\circ})\)

\(\tan(-690^{\circ})=\tan(30^{\circ})=\frac{\sin(30^{\circ})}{\cos(30^{\circ})}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\).

Step5: Calculate \(\cot(-690^{\circ})\)

\(\cot(-690^{\circ})=\cot(30^{\circ})=\frac{\cos(30^{\circ})}{\sin(30^{\circ})}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\).

Step6: Calculate \(\csc(-690^{\circ})\)

\(\csc(-690^{\circ})=\csc(30^{\circ})=\frac{1}{\sin(30^{\circ})}=2\).

Step7: Calculate \(\sec(-690^{\circ})\)

\(\sec(-690^{\circ})=\sec(30^{\circ})=\frac{1}{\cos(30^{\circ})}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\).

Answer:

\(\sin(-690^{\circ})=\frac{1}{2}\)
\(\cos(-690^{\circ})=\frac{\sqrt{3}}{2}\)
\(\tan(-690^{\circ})=\frac{\sqrt{3}}{3}\)
\(\cot(-690^{\circ})=\sqrt{3}\)
\(\csc(-690^{\circ}) = 2\)
\(\sec(-690^{\circ})=\frac{2\sqrt{3}}{3}\)