Roots, signs and abs

by Jhenrique
Tags: roots, signs
 P: 686 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/Hyperb...chnungstabelle and here is a full trigonometric list for comparation: https://es.wikipedia.org/wiki/Identi...s_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?
P: 477
From the Wiki that you linked ... immediately above the table that is apparently in question:

 De estas dos identidades, se puede extrapolar la siguiente tabla. Sin embargo, nótese que estas ecuaciones de conversión pueden devolver el signo incorrecto (+ ó −).
Mentor
P: 21,314
 Quote by Jhenrique By pythagorean identity, ##\sin(x)^2 + \cos(x)^2 = 1##, so ##\sin(x) = \sqrt{1 - \cos(x)^2}##;
No. You omitted the ##\pm##.
##\sin(x) = \pm \sqrt{1 - \cos(x)^2}##
 Quote by Jhenrique also, ##\sinh(x)^2 - \cosh(x)^2 = - 1##, therefore ##\sinh(x) = \sqrt{\cosh(x)^2 - 1}##.
Again, no, same problem as above.
##\sinh(x) = \pm \sqrt{\cosh(x)^2 - 1}##
 Quote by Jhenrique;4708956 Happens that the last equation is incorrect, here is a full list of the correct forms for the hyperbolics: [url https://de.wikipedia.org/wiki/Hyperbelfunktion#Umrechnungstabelle[/url] and here is a full trigonometric list for comparation: https://es.wikipedia.org/wiki/Identi...s_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?

 P: 686 Roots, signs and abs Yeah, I like of omit +/- because, by definition, a root square have 2 roots...
Mentor
P: 21,314
 Quote by Jhenrique Yeah, I like of omit +/- because, by definition, a root square have 2 roots...
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.
 P: 686 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?
P: 834
 Quote by Jhenrique 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?
On the reals there are only two roots: ##\sqrt[6]{x}## and ##-\sqrt[6]{x}##.
Jhenrique, you have reals and complex numbers mixed up.
HW Helper
P: 3,540
 Quote by Jhenrique 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?
If you take the square root of both sides of

$$y^6=x$$

you get

$$y^3=\pm\sqrt{x}$$
Mentor
P: 21,314
 Quote by Jhenrique 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?
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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