How do you evaluate #Sin(pi/2) + 6 cos(pi/3)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Noah G Nov 8, 2016 #=4# Explanation: #sin(pi/2) = 1# #cos(pi/3) = 1/2# #1 + 6(1/2) = 1 + 3 = 4# Hopefully this helps! 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 2049 views around the world You can reuse this answer Creative Commons License