How do you find exact value of #cos (pi/12)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Bdub Mar 30, 2016 #cos (pi/12)=sqrt6/4+sqrt2/4# Explanation: #cos (pi/12)=cos(pi/4 -pi/6)# #cos(A-B)=cosAcosB+sinAsinB# #cos(pi/4 -pi/6)=cos (pi/4)cos(pi/6)+sin(pi/4)sin(pi/6)# #cos(pi/4 -pi/6)=sqrt2/2*sqrt3/2+sqrt2/2*1/2=sqrt6/4+sqrt2/4# #cos (pi/12)=sqrt6/4+sqrt2/4# 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 89816 views around the world You can reuse this answer Creative Commons License