How do you calculate #arccos(- sqrt3/2)#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Massimiliano May 1, 2015 Since the range of the function #y=arccosx# is #[0,pi]# and the value negative #-sqrt3/2#, the angle is in the second quadrant. So: #arccos(-sqrt3/2)=5/6pi#. 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 16862 views around the world You can reuse this answer Creative Commons License