Roots, signs and abs

  1. By pythagorean identity, ##\sin(x)^2 + \cos(x)^2 = 1##, so ##\sin(x) = \sqrt{1 - \cos(x)^2}##; also, ##\sinh(x)^2 - \cosh(x)^2 = - 1##, therefore ##\sinh(x) = \sqrt{\cosh(x)^2 - 1}##.

    Happens that the last equation is incorrect, here is a full list of the correct forms for the hyperbolics:
    https://de.wikipedia.org/wiki/Hyperbelfunktion#Umrechnungstabelle and here is a full trigonometric list for comparation: https://es.wikipedia.org/wiki/Identidades_trigonométricas#Relaciones_b.C3.A1sicas.

    So, why the 'normal' trigonometrics no needs of completary functions, like Abs and Sgn, and the hyperbolic trigonometrics needs in some case?
     
  2. jcsd
  3. From the Wiki that you linked ... immediately above the table that is apparently in question:

     
  4. Mark44

    Staff: Mentor

    No. You omitted the ##\pm##.
    ##\sin(x) = \pm \sqrt{1 - \cos(x)^2}##
    Again, no, same problem as above.
    ##\sinh(x) = \pm \sqrt{\cosh(x)^2 - 1}##
     
  5. Yeah, I like of omit +/- because, by definition, a root square have 2 roots...
     
  6. Mark44

    Staff: Mentor

    No, that's not the definition. The square root of a positive real number has one value, not two.


    It's true that real numbers have two square roots -- one positive and one negative -- but the expression ##\sqrt{x}## represents the principal square root of x, a positive real number that when multiplied by itself yields x.

    If a square root represented two values, there would be no need to write ##\pm## in the quadratic formula:
    $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

    When you start with sin2(x) + cos2(x) = 1 and solve for sin(x), you need ##\pm## in there, otherwise you are getting only the positive value.
     
  7. And if you have ##x=y^6## ? You'll write ##\pm\sqrt{\pm\sqrt{\pm\sqrt{x}}}## ? Not is better let that the ##\sqrt[n]{x}## represents the n roots?
     
  8. pwsnafu

    pwsnafu 938
    Science Advisor

    On the reals there are only two roots: ##\sqrt[6]{x}## and ##-\sqrt[6]{x}##.
    Jhenrique, you have reals and complex numbers mixed up.
     
  9. Mentallic

    Mentallic 3,763
    Homework Helper

    If you take the square root of both sides of

    [tex]y^6=x[/tex]

    you get

    [tex]y^3=\pm\sqrt{x}[/tex]
     
  10. Mark44

    Staff: Mentor

    Let's make it simple.
    ##y^6 = 64##
    ##\Rightarrow y = \pm \sqrt[6]{64} = \pm 2##

    As it turns out, there are four other sixth roots of 64, but they are all complex. The only real sixth roots of 64 are 2 and -2.
     
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