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Roots, signs and abs 
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#1
Apr314, 07:46 PM

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? 


#2
Apr314, 09:23 PM

P: 425

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



#3
Apr314, 11:47 PM

Mentor
P: 21,216

##\sin(x) = \pm \sqrt{1  \cos(x)^2}## ##\sinh(x) = \pm \sqrt{\cosh(x)^2  1}## 


#4
Apr414, 12:40 AM

P: 686

Roots, signs and abs
Yeah, I like of omit +/ because, by definition, a root square have 2 roots...



#5
Apr414, 12:59 AM

Mentor
P: 21,216

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 sin^{2}(x) + cos^{2}(x) = 1 and solve for sin(x), you need ##\pm## in there, otherwise you are getting only the positive value. 


#6
Apr414, 03:44 AM

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?



#7
Apr414, 05:31 AM

Sci Advisor
P: 820

Jhenrique, you have reals and complex numbers mixed up. 


#8
Apr414, 05:36 AM

HW Helper
P: 3,511

[tex]y^6=x[/tex] you get [tex]y^3=\pm\sqrt{x}[/tex] 


#9
Apr414, 10:26 AM

Mentor
P: 21,216

##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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