How do you find the exact value of #sin^-1(-sqrt3/2)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Vinícius Ferraz Oct 6, 2016 #{- pi / 3, - (2 pi) / 3} + 2 pi cdot ZZ# Explanation: #sin^-1 (-sqrt 3 / 2) = x# #-sqrt 3 / 2 = sin x# #x_1 = -60º or x_2 = - 180º + 60º# if #x_i# is a solution, then #x_i + 2 k pi# are also solutions. #k in ZZ# #- pi / 3 + 2 k pi# or #- pi + pi / 3 + 2 k pi# Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove #\csc \theta \times \tan \theta = \sec \theta#? How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#? How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? How do you verify the identity: #-cotx =(sin3x+sinx)/(cos3x-cosx)#? How do you prove that #(tanx+cosx)/(1+sinx)=secx#? How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#? See all questions in Proving Identities Impact of this question 1798 views around the world You can reuse this answer Creative Commons License