# Roots, signs and abs

1. Apr 3, 2014

### Jhenrique

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. Apr 3, 2014

### gopher_p

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

3. Apr 3, 2014

### 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. Apr 4, 2014

### Jhenrique

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

5. Apr 4, 2014

### 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. Apr 4, 2014

### 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?

7. Apr 4, 2014

### pwsnafu

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. Apr 4, 2014

### Mentallic

If you take the square root of both sides of

$$y^6=x$$

you get

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

9. Apr 4, 2014

### 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.