How do you evaluate #sin(11(pi)/2) #? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Karthik G Feb 1, 2016 Explanation is given below Explanation: #sin((11pi)/2)# #=sin(6pi-pi/2)# Note #color(magenta)(sin(2npi - theta) = -sin(theta)# and #color(magenta)(sin(2npi + theta) = sin(theta)# #=-sin(pi/2)# #=-1# 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 3317 views around the world You can reuse this answer Creative Commons License