How do you evaluate #cos((11pi) / 12)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N. Mar 20, 2016 #sqrt(2 + sqrt3)/2# Explanation: On the trig unit circle, #cos((11pi)/12) = cos (pi/12#). Evaluate #cos (pi/12)# bu applying the identity: #cos 2a = 2cos^2 a - 1# #cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1# #2cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/2# #cos^2 (pi/12) = (2 + sqrt3)/4# #cos (pi/12) = sqrt(2 + sqrt3)/2# (#cos (pi/12)# is positive) 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 1551 views around the world You can reuse this answer Creative Commons License