How do you evaluate #cos((5*pi) / 6) - tan((-7*pi) / 4) + sin((13*pi) / 3)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer P dilip_k Oct 27, 2016 #cos((5pi)/6)-tan(-(7pi)/4)+sin((13pi)/3)# #=cos(pi-pi/6)+tan(2pi-pi/4)+sin(4pi+pi/3)# #=-cos(pi/6)-tan(pi/4)+sin(pi/3)# #=-sqrt3/2-1+sqrt3/2# #=-1# 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 1970 views around the world You can reuse this answer Creative Commons License