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P. Moller - Maersk. com 42 The Elementary Functions Calculus 1a In this case we have � π π� , y = Arcsin x ∈ − , 2 2 sin(Arcsin x) = x, (45) 1 d , Arcsin x = √ dx 1 − x2 for x ∈ [−1, 1], for x ∈ [−1, 1], for x ∈ ] − 1, 1[. 4 2 –4 –2 0 2 4 –2 –4 Figure 19: The graphs of y = Arctan x, x ∈ R, and y = tan x, x ∈ ] − π/2, π/2[ (dotted with dotted asymptotes). The function Arctan x. Among many possibilities we choose the restriction � π π� . y = tan x ∈ R for x ∈ − , 2 2 In this case we have � π π� , y = arctan x ∈ − , 2 2 tan(Arctan x) = x, (46) 1 d Arctan x = , 1 + x2 dx for x ∈ R, for x ∈ R, for x ∈ R.
Other functional relations can be found in the appendix. Connection with the complex exponential function. 6 by ei x = cos x + i · sin x. One example is the calculation ei(x+y) = cos(x + y) + i · sin(x + y) = ei x · ei y = {cos x + i sin x} · {cos y + i sin y} = cos x · cos y − sin x · sin y + i · {sin x · cos y + cos x · sin y}. com 30 The Elementary Functions Calculus 1a By a splitting into the real and imaginary parts we obtain again the addition formulæ, cos(x + y) = cos x · cos y − sin x · sin y, sin(x + y) = sin x · cos y + cos x · sin y.
The derivation of the inverse functions of the other hyperbolic functions is left to the reader. 5 Figure 13: The graphs of y = cosh x, x ≥ 0, and y = Arcosh x, x ≥ 1. The inverse function of y = cosh x. When we consider the graph we conclude that the most natural restriction is given by cosh : [0, +∞[ → [1, +∞[. The task is now after interchanging the letters to solve the equation x = cosh y, x ≥ 1 and y ≥ 0. Since x = cosh y = � � 1 � 1� y 2 e + e−y = y (ey ) + 1 , 2e 2 we get by a rearrangement, 2 (ey ) − 2x · ey + 1 = 0, x ≥ 1 and y ≥ 0.