How do you evaluate #sin(pi/8) #? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer A. S. Adikesavan Feb 18, 2016 #sqrt#((#sqrt#2# -# 1)/(2#sqrt#2)) Explanation: cos#pi/4# = 1 #-# 2 sin#pi/8# sin#pi/8#. cos #pi/4# = 1/#sqrt#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 8689 views around the world You can reuse this answer Creative Commons License