How do you evaluate #cos(pi/3)cos((5pi)/12)-sin(pi/3)sin((5pi)12)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Ratnaker Mehta Jul 11, 2016 #-1/sqrt2.# Explanation: Recall that #cos(A+B)=cosAcosB-sinAsinB# So, given Exp. #=cos(pi/3+5pi/12)=cos(9pi/12)=cos(3pi/4)=cos (pi-pi/4)=-cos(pi/4)=-1/sqrt2.# 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 1735 views around the world You can reuse this answer Creative Commons License