How do you solve the inverse trig function #cos^-1 (-sqrt2/2)#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Nghi N · Hriman Mar 7, 2018 # (3pi)/4# Explanation: Find #arccos (- sqrt2/2)#. #cos x = - sqrt2/2# Trig Table and unit circle give 1 solution arc: #x = (3pi)/4# 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 #sin(sin^-1 (1/3))#? How do you solve the inverse trig function #arcsin (sin 5pi/6)#? See all questions in Inverse Trigonometric Properties Impact of this question 5873 views around the world You can reuse this answer Creative Commons License