How do you evaluate #cot ((17pi)/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Konstantinos Michailidis Mar 18, 2016 It is #cot(17*pi/6)=cot([12pi+5pi]/6)=cot(2pi+5pi/6)=cot(5pi/6)= =cot(pi-pi/6)=-cot(pi/6)=-sqrt3# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 5717 views around the world You can reuse this answer Creative Commons License