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Differentiation of inverse trig

  1. Feb 26, 2014 #1
    Say we want to differentiate [itex]\arcsin x[/itex]. To do this we put [tex]y=\arcsin x.[/tex] Then [tex]x=\sin y \implies \frac{dx}{dy}= \cos y.[/tex] Then we use the relation [tex]\sin^2 y + \cos^2 y = 1 \implies \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}.[/tex] Therefore [tex]\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}.[/tex]

    My question is that when we use [tex]\sin^2 y + \cos^2 y = 1,[/tex] how do we know that [tex]\cos y = \sqrt{1 - \sin^2 y}[/tex] rather than [tex]\cos y = -\sqrt{1 - \sin^2 y}[/tex]?
     
  2. jcsd
  3. Feb 26, 2014 #2
    Because arcsine, by convention, always returns angles in the range [itex][-\pi/2, \pi/2][/itex]. This guarantees that [itex]y = \arcsin x[/itex] is in that range, so [itex]\cos y[/itex] has to be positive. Therefore, you know it must be the positive square root, instead of the negative one.
     
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