How do you evaluate # cos (pi/2) cos(pi/7) + sin(pi/2) sin(pi/7)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer P dilip_k Oct 16, 2016 Inserting #cos(pi/2)=0 and sin(pi/2)=1# in the given expression we get #cos(pi/2)cos(pi/7)+sin(pi/2)sin(pi/7)# #=0xxcos(pi/7)+1xxsin(pi/7)# #=sin(pi/7)=0.434# 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 2765 views around the world You can reuse this answer Creative Commons License