How do you evaluate #Cos^-1 (cos -pi/6)#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer P dilip_k Mar 27, 2016 The pricipal value of #theta =cos^-1x# should #0<=theta<=pi# So #cos^-1(cos(-pi/6))=cos^-1(sqrt3/2)=pi/6# 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 19008 views around the world You can reuse this answer Creative Commons License