How do you find the value of #cos(pi/8)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Konstantinos Michailidis Jun 19, 2016 The half angle formula for cosine is #cos x= 2 cos^2 (x/2) -1#. Thus cos #pi/4 = 2 cos^2 (pi/8) -1#, #cos^2 (pi/8)= ( 1+ cos (pi/4))/2# =#( 1+sqrt2)/(2sqrt2)#= #(2+sqrt2)/4# cos #pi/8#= #sqrt (2+sqrt2) /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 6947 views around the world You can reuse this answer Creative Commons License