How do you solve the inverse trig function cos (arccos 3pi)cos(arccos3π)? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Luca F. Jun 11, 2015 It does not exist. Explanation: y=arccos(x)y=arccos(x) is a function defined in the interval [-1,1][−1,1] So it does not exist for the value 3pi>13π>1. arccos(3pi)arccos(3π) does not make any sense! Answer link Related questions How do you use the properties of inverse trigonometric functions to evaluate tan(arcsin (0.31))tan(arcsin(0.31))? What is \sin ( sin^{-1} frac{sqrt{2}}{2})sin(sin−1√22)? How do you find the exact value of \cos(tan^{-1}sqrt{3})cos(tan−1√3)? How do you evaluate \sec^{-1} \sqrt{2} sec−1√2? How do you find cos( cot^{-1} sqrt{3} )cos(cot−1√3) without a calculator? How do you rewrite sec^2 (tan^{-1} x)sec2(tan−1x) 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)sin−1(0.1)? How do you solve the inverse trig function cos^-1 (-sqrt2/2)cos−1(−√22)? How do you solve the inverse trig function sin(sin^-1 (1/3))sin(sin−1(13))? See all questions in Inverse Trigonometric Properties Impact of this question 4943 views around the world You can reuse this answer Creative Commons License