How do you evaluate sin(arccos(-1/3))? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N. Oct 28, 2015 Evaluate sin (arccos (-1/3) Ans: #+- (2sqrt2)/3# Explanation: #cos x = -1/3# --> Radius of trig circle R = 3 #sin^2 x = R^2 - cos^2 x = 9 - 1/9 = 8/9# --> #sin x = +- (2sqrt2)/3# #sin (arccos (-1/3)) = +- (2sqrt2)/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 4828 views around the world You can reuse this answer Creative Commons License