How do you evaluate #cos(-pi/6) cos (pi/4)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N. Jun 17, 2016 #sqrt6/4# Explanation: Trig table --> #cos (pi/4) = sqrt2/2# #cos (-pi/6) = cos (pi/6) = sqrt3/2# #P = cos (pi/4).cos (-pi/6) = (sqrt2/2)(sqrt3/2) = sqrt6/4# 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 2852 views around the world You can reuse this answer Creative Commons License