Inverse

1 1 z = sec ϑ The inverse sine . The sine function restricted to [ π, π ] is one-to-one, and its inverse on this 3 − 2 2 ϑ = π ϑ = arcsec z interval is called the arcsine (arcsin) function. The domain of arcsin is [ 1, 1] and the range of arcsin is 2 1 1 1 −1 [ π, π ]. Below is a graph of y = sin ϑ, with the the part over [ π, π ] emphasized, and the graph − 2 2 − 2 2 of ϑ y. π 1 = arcsin ϑ = 2 π y = sin ϑ ϑ = arcsin y 1 π π 2 π − 1 ϑ 1 1 z −

1 π 1 π ϑ 1 1 y − 2 2 − 1 − 1 1 3 π ϑ = 2 π ϑ = 2 π − 2 By definition, By definition, 1 1 1 3 ϑ = arcsin y means that sin ϑ = y and π 6 ϑ 6 π. ϑ = arcsec z means that sec ϑ = t and 0 6 ϑ< 2 π or π 6 ϑ< 2 π. − 2 2 Differentiating the equation on the right implicitly with respect to y, gives Again, differentiating the on the right implicitly with respect to z, and using the restriction in ϑ,

dϑ dϑ 1 1 1 one computes the cos ϑ = 1, or = , provided π<ϑ< π. dy dy cos ϑ − 2 2 d 1 (arcsec z)= 2 , for z > 1. 1 1 2 2 dz z√z 1 | | Since cos ϑ> 0 on ( 2 π, 2 π ), it follows that cos ϑ = p1 sin ϑ = p1 y . Therefore, − − − − The other inverse trigonometric functions. The remaining trigonometric functions can be defined via the d 1 (arcsin y)= , for 1

π π ϑ t − x u 1 1 1 ϑ = π − − 2 1 arccsc w = π arcsec w 2 − ϑ = arccsc w 1 1 ϑ = 2 π ϑ = 2 π − 1 By definition, 2 π 1 1 ϑ = arctan t means that tan ϑ = t and π<ϑ< π. − 2 2 As with arcsin, one can use the equation on the right, and the restriction on ϑ, to compute the derivative of 1 1 w − 1 the arctangent function: 2 π d 1 − (arctan t)= , for t R. dt 1+ t2 ∈ 1 3 ϑ = π The inverse secant function. The secant function restricted to [ 0, π ) [ π, π ) is one-to-one, and its in- − 2 ∪ 2 verse on this interval is called the arcsecant (arcsec) function. The domain of arcsec is ( , 1] [ 1, ) 1 3 −∞ − ∪ ∞ What is the domain and range of each of these functions? You should try to derive the formulæ for the and the range of arcsec is [ 0, 2 π ) [ π, 2 π ). Below is a graph of z = sec ϑ, with the the part over 1 3 ∪ of the inverse trigonometric functions besides arcsin. (Note that computing the derivatives of the [ 0, π ) [ π, π ). emphasized, and the graph of ϑ = arcsec z. 2 ∪ 2 last three will be easier.) Finally, identify any vertical tangents, and/or horizontal asymptotes of the graphs of each of the inverse trigonometric functions.