How do you evaluate cos(arcsin(-1/2)+arctan(2/3))? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 1 Answer A. S. Adikesavan Dec 19, 2016 sqrt13/26(3sqrt3+2) =0.9979, nearly. Explanation: Let a = arc sin (-1/2) = -pi/6 and b = arc tan (2/3) in Q_1, giving tan b = 2/3, sin b = 2/sqrt 13 and cos b =3/sqrt 13. The given expression is cos (a+b) = cos (-pi/6+b)=cos (-pi/6)cosb-sin(-pi/6)sin b =sqrt 3/2(3/sqrt 13)+1/2(2/sqrt 13) =sqrt13/26(3sqrt3+2) =0.9979, nearly. Answer link Related questions What are the Basic Inverse Trigonometric Functions? How do you use inverse trig functions to find angles? How do you use inverse trigonometric functions to find the solutions of the equation that are in... How do you use inverse trig functions to solve equations? How do you evalute sin^-1 (-sqrt(3)/2)? How do you evalute tan^-1 (-sqrt(3))? How do you find the inverse of f(x) = \frac{1}{x-5} algebraically? How do you find the inverse of f(x) = 5 sin^{-1}( frac{2}{x-3} )? What is tan(arctan 10)? How do you find the arcsin(sin((7pi)/6))? See all questions in Basic Inverse Trigonometric Functions Impact of this question 4119 views around the world You can reuse this answer Creative Commons License