What is #(11pi)/6# in degrees and hence find #cos((11pi)/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria Mar 28, 2017 #cos((11pi)/6)=sqrt3/2# Explanation: As #pi# radians are equal to #180^@# #1# radians are equal to #180^@/pi# #(11pi)/6# radians are #(cancel(180)30^@)/cancelpixx(11cancelpi)/(cancel6^1)# = #30^@xx11=330^@# and #cos((11pi)/6)=cos330^@# = #cos(360^@-30^@)# = #cos(-30^@)# = #cos30^@# = #sqrt3/2# 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 2215 views around the world You can reuse this answer Creative Commons License