How do you calculate Arccos (sin (pi/6))? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Nghi N. Jul 26, 2015 Find arccos(sin (pi/6)) Ans: +- pi/3 Explanation: sin (pi/6) = 1/2 therefor arccos (sin(pi/6)) = arccos (1/2) On the trig unit circle, cos x = 1/2 --> arc x = +- pi/3 Answer link Related questions How do you use the properties of inverse trigonometric functions to evaluate tan(arcsin (0.31))? What is \sin ( sin^{-1} frac{sqrt{2}}{2})? How do you find the exact value of \cos(tan^{-1}sqrt{3})? How do you evaluate \sec^{-1} \sqrt{2} ? How do you find cos( cot^{-1} sqrt{3} ) without a calculator? How do you rewrite sec^2 (tan^{-1} x) in terms of x? How do you use the inverse trigonometric properties to rewrite expressions in terms of x? How do you calculate sin^-1(0.1)? How do you solve the inverse trig function cos^-1 (-sqrt2/2)? How do you solve the inverse trig function sin(sin^-1 (1/3))? See all questions in Inverse Trigonometric Properties Impact of this question 16367 views around the world You can reuse this answer Creative Commons License