How do you solve the inverse trig function #sin(sin^-1 (1/3))#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Alan P. May 25, 2015 #sin^-1(1/3)# or #arcsin(1/3)# is the angle #theta# for which #sin(theta) = 1/3# Therefore #sin(sin^-1(1/3))# #= sin(theta)# for the value of #theta# for which #sin(theta) = 1/3# That is #sin(sin^-1(1/3)) = 1/3# 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 #arcsin (sin 5pi/6)#? See all questions in Inverse Trigonometric Properties Impact of this question 4623 views around the world You can reuse this answer Creative Commons License