How do you evaluate #sin ((-5pi)/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria Oct 28, 2016 #sin(-(5pi)/6)=-1/2# Explanation: As #sin(-theta)=-sintheta# and #sin(pi-theta)=sintheta# #sin(-(5pi)/6)=-sin((5pi)/6)# = #-sin((6pi-pi)/6)=-sin(pi-pi/6)# = #-sin(pi/6)# = #-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 8102 views around the world You can reuse this answer Creative Commons License