How do you evaluate arc cos(-1/2)arccos(−12)? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 1 Answer Shwetank Mauria Nov 26, 2016 arccos(-1/2)=(2pi)/3arccos(−12)=2π3 Explanation: arccosxarccosx is an angle whose cosine ratio is -1/2−12. Now as cos(pi/3)=1/2cos(π3)=12 and cos (pi-A)=-cosAcos(π−A)=−cosA cos(pi-pi/3)cos(π−π3) i.e. cos((2pi)/3)=-coe(pi/3)=-1/2cos(2π3)=−coe(π3)=−12 As cos((2pi)/3)=-1/2cos(2π3)=−12, we have arccos(-1/2)=(2pi)/3arccos(−12)=2π3 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)sin−1(−√32)? How do you evalute tan^-1 (-sqrt(3))tan−1(−√3)? How do you find the inverse of f(x) = \frac{1}{x-5}f(x)=1x−5 algebraically? How do you find the inverse of f(x) = 5 sin^{-1}( frac{2}{x-3} )f(x)=5sin−1(2x−3)? What is tan(arctan 10)? How do you find the arcsin(sin((7pi)/6))arcsin(sin(7π6))? See all questions in Basic Inverse Trigonometric Functions Impact of this question 8292 views around the world You can reuse this answer Creative Commons License