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Trigonometric equation

  1. Apr 13, 2014 #1
    Is correct the following procediment?

    ## A \sin (\omega t) = A \sin (\phi) \to \phi= \sin^{-1} \sin (\omega t )##

    Is correct to say that ## \phi = \omega t## is oscillatory in this case ??
     
  2. jcsd
  3. Apr 13, 2014 #2

    Mark44

    Staff: Mentor

    You don't need inverse trig functions here. We can ignore A, assuming that it is a nonzero constant.
    If sin(ωt) = sin(θ), then θ = ωt + k(##2\pi##), where k is an integer.
     
  4. Apr 13, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    But you certainly cannot "say that [itex]\phi= \omega t[/itex] is oscillatory"! It is the function [itex]sin(\phi)[/itex] that is oscillatory, not just the argument, [itex]\phi[itex].
     
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