Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook