How do you evaluate #sin (pi / 12) * cos (3 pi / 4) - cos (pi / 12) * sin (3 pi / 4)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer RedRobin9688 Jun 27, 2018 #-sqrt3/2# Explanation: Recall #sin(A-B)=sinAcosB-cosAsinB# #sin(pi/12)cos((3pi)/4)-cos(pi/12)sin((3pi)/4)=sin(pi/12-(3pi)/4)# #sin(pi/12-(9pi)/12)# Recall #sin(-x)=-sin(x)# #sin(-(2pi)/3)=-sin((2pi)/3)# #-sqrt3/2# 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 4062 views around the world You can reuse this answer Creative Commons License