How do you evaluate sine, cosine, tangent of #(10pi)/3# without using a calculator? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Andrea S. Dec 28, 2016 #sin((10pi)/3)= -sqrt(3)/2# #cos((10pi)/3)= -1/2# #tan((10pi)/3)= sqrt(3)# Explanation: You consider that: #(10pi)/3 = 2pi+pi+pi/3#, so that: #sin((10pi)/3)= sin(2pi+pi+pi/3) =sin(pi+pi/3) = sinpicos(pi/3)+cospisin(pi/3) = -sin(pi/3)= -sqrt(3)/2# #cos((10pi)/3)= cos(2pi+pi+pi/3) =cos(pi+pi/3) = cospicos(pi/3)-sinpisin(pi/3) = -cos(pi/3)= -1/2# #tan((10pi)/3)= sin((10pi)/3)/(cos((10pi)/3))=(-sqrt(3)/2)/(-1/2) = sqrt(3)# 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 15828 views around the world You can reuse this answer Creative Commons License