# Roots, signs and abs

1. ### Jhenrique

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/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. ### gopher_p

575
From the Wiki that you linked ... immediately above the table that is apparently in question:

### 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}##

4. ### Jhenrique

686
Yeah, I like of omit +/- because, by definition, a root square have 2 roots...

### 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.

6. ### Jhenrique

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?

7. ### pwsnafu

933
On the reals there are only two roots: ##\sqrt[6]{x}## and ##-\sqrt[6]{x}##.
Jhenrique, you have reals and complex numbers mixed up.

8. ### Mentallic

3,745
If you take the square root of both sides of

$$y^6=x$$

you get

$$y^3=\pm\sqrt{x}$$

### 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.