How do you solve the inverse trig function #cos (arccos 3pi)#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Luca F. Jun 11, 2015 It does not exist. Explanation: #y=arccos(x)# is a function defined in the interval #[-1,1]# So it does not exist for the value #3pi>1#. #arccos(3pi)# 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))#? 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 4605 views around the world You can reuse this answer Creative Commons License