Proofs w/ hyperbolic fcns: sinh^2(x) = 1/(1 + coth^2(x)) and sinh(x/2) = sgn(x) sqrt{(cosh(x) - 1)/2}

dr.trovacek

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Can't prove these, need help.

1.
\(\displaystyle \sinh^2x = \frac {1}{1 + \coth^2 x}\)

So I took the right side: \(\displaystyle \left ( 1 + \frac{\cosh^2 x}{\sinh^2 x} \right )^{-1} = \frac{\sinh^2 x}{\cosh^2 x + \sinh^2 x}\)

The denominator is not equal to 1, so where did I go wrong?

Tried to check it with mathway.com, but it tells me that expression is not true

2.
\(\displaystyle \sinh \frac { x } { 2 } = \operatorname { sgn } ( x ) \cdot \sqrt { \frac { \cosh x - 1 } { 2 } }\)

Never solved an equation with \(\displaystyle \operatorname { sgn }\) in it, so I assume that may be the problem. I just squared both sides of the equation and assumed \(\displaystyle \operatorname { sgnx} = \pm 1\) but it doesn't work .
 

tkhunny

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On the first, Have you considered that 1) the statement is written incorrectly, or 2) you might be more successful Disproving it?

On the second, why not just solve it thrice? 1) x > 0 and sgn(x) = 1, or 2) x < 0 and sgn(x) = -1, or 3) x = 0 and sgn(0) = 0
 

pka

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Can't prove these, need help.
1. \(\displaystyle \sinh^2x = \frac {1}{1 + \coth^2 x}\)
So I took the right side: \(\displaystyle \left ( 1 + \frac{\cosh^2 x}{\sinh^2 x} \right )^{-1} = \frac{\sinh^2 x}{\cosh^2 x + \sinh^2 x}\)
The denominator is not equal to 1, so where did I go wrong?
Tried to check it with mathway.com, but it tells me that expression is not true
When working with hyperbolics I suggest using the definitions.
\(\displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}\text{ and }\cosh(x)=\frac{e^x+e^{-x}}{2}.\)
So that \(\displaystyle \sinh^2(x)=\frac{e^{2x}-2+e^{-2x}}{4}\)
Because \(\displaystyle \cosh^2(x)=\sinh^2(x)+1\) then \(\displaystyle \cosh^2(x)+\sinh^2(x)=2\sinh^2(x)+1\)
Although I have not done it, I too doubt the equality.
 
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dr.trovacek

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On the first, Have you considered that 1) the statement is written incorrectly, or 2) you might be more successful Disproving it?

1) I did consider this, but I'm not sure. That is one of the reasons I asked for help here. This is from my fourth grade high school textbook and since I work with hyperbolic functions for the first time I thought I could easily be mistaken.

2) I'm not sure if I would know hot to disprove it, at least not in some mathematically rigorous way. Could you give me some guidelines or show how you would do it?

On the second, why not just solve it thrice? 1) x > 0 and sgn(x) = 1, or 2) x < 0 and sgn(x) = -1, or 3) x = 0 and sgn(0) = 0
I managed to solve the second problem earlier today. I was constantly trying to solve it algebraically. I did just what you have now said and managed to get it right - feeling stupid :)
 

dr.trovacek

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When working with hyperbolics I suggest using the definitions.
\(\displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}\text{ and }\cosh(x)=\frac{e^x+e^{-x}}{2}.\)
So that \(\displaystyle \sinh^2(x)=\frac{e^{2x}-2+e^{-2x}}{4}\)
Because \(\displaystyle \cosh^2(x)=\sinh^2(x)+1\) then \(\displaystyle \cosh^2(x)+\sinh^2(x)=2\sinh^2(x)+1\)
Although I have not done it, I too doubt the equality.

Yeah I tried a couple of things using the definitions but still unsuccessfully.
 
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tkhunny

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Yeah I tried a couple of things using the definitions but I still unsuccessfully.
Proof AGAINST requires only ONE example. Try anything EXCEPT x = 0.
 

dr.trovacek

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Thank you both
 
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