Question #d532a Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer kamil9234 Apr 6, 2017 #sin((8pi)/3)=sqrt(3)/3# #cos((8pi)/3)=-1/2# Explanation: #(8pi)/3=2pi and (2pi)/3# #2pi# is full period and you can remove it. SO, You have: #sin((2pi)/3)# You can use this: #sin(pi-alpha)=sinalpha# #cos(pi-alpha)=-cosalpha# #sin((2pi)/3)=sin(pi-pi/3)=sin(pi/3)=sqrt(3)/3# #cos((2pi)/3)=cos(pi-pi/3)=-cos(pi/3)=-1/2# 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 1276 views around the world You can reuse this answer Creative Commons License