How do you find the range and domain of #sin(arctan x)#? Trigonometry Inverse Trigonometric Functions Graphing Inverse Trigonometric Functions 1 Answer Shwetank Mauria Jun 11, 2017 Range for #sin(arctanx)# is #[-1,1]#, and domain is #(-oo,oo)#. Explanation: #arctanx# is always from #-pi/2# to #pi/2#, but here #x# can take any value from #(-oo.oo)#, which is domain for #sin(arctanx)#. But #sin(arctanx)# can take values only from #[-1,1]# Hence range for #sin(arctanx)# is #[-1,1]#. Answer link Related questions Question #b4e98 How do you graph inverse trigonometric functions? What is the domain and range of inverse trigonometric functions? How do you apply the domain, range, and quadrants to evaluate inverse trigonometric functions? Can the values of the special angles of the unit circle be applied to the inverse trigonometric... How do you graph #y = 2\sin^{-1}(2x)#? What is the domain and range for #y = xcos^-1[x]#? What is the domain and range for #y = 6sin^-1(4x)#? How do you find the domain and range for #y = 5arcsin(2cos(3x))#? How do you graph #y = Arctan(x/3) #? See all questions in Graphing Inverse Trigonometric Functions Impact of this question 6949 views around the world You can reuse this answer Creative Commons License