How do you evaluate #cos ((53 pi) / 6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer José F. Mar 22, 2016 #-sqrt(3)/2# Explanation: #cos((53pi)6)=cos(((48+5)pi)/6)=cos (4pi+5pi/6)=cos (5pi/6)# #cos (5pi/6)=cos (pi-pi/6)=-cos(pi/6)=-sqrt(3)/2# 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 6331 views around the world You can reuse this answer Creative Commons License