January 1st, 2017, 06:46 AM  #1 
Newbie Joined: Oct 2015 From: huddersfield Posts: 16 Thanks: 1  hyperbolic identities
Prove the hyperbolic identities coth x ≡ cosech 2x + tanh x 
January 1st, 2017, 08:43 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,691 Thanks: 1350  
January 1st, 2017, 06:45 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,692 Thanks: 976 Math Focus: Elementary mathematics and beyond 
Use the identity $\sinh2x=2\sinh x\cosh x$

January 4th, 2017, 10:55 AM  #4 
Newbie Joined: Oct 2015 From: huddersfield Posts: 16 Thanks: 1  
January 4th, 2017, 11:34 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,691 Thanks: 1350 
$\coth{x} = 2\text{csch}(2x) + \tanh{x}$ $\coth{x} = \dfrac{2}{\sinh(2x)} + \tanh{x}$ use greg1313's recommendation ... $\coth{x} = \dfrac{2}{2\sinh{x}\cosh{x}} + \tanh{x}$ $\coth{x} = \dfrac{1}{\sinh{x}\cosh{x}} + \dfrac{\sinh{x}}{\cosh{x}}$ $\coth{x} = \dfrac{1}{\sinh{x}\cosh{x}} + \dfrac{\sinh^2{x}}{\sinh{x}\cosh{x}}$ $\coth{x} = \dfrac{1+ \sinh^2{x}}{\sinh{x}\cosh{x}}$ recall $1 = \cosh^2{x}\sinh^2{x}$ ... can you finish from here? 
January 4th, 2017, 11:46 AM  #6 
Newbie Joined: Oct 2015 From: huddersfield Posts: 16 Thanks: 1 
skeeter, you seem to have a very good understanding of this topic, whereas I believe mine is not so good; where is the best place to gain a better understanding of this subject? I can't seem to grasp the next stages after your answers.
Last edited by skipjack; January 4th, 2017 at 12:08 PM. 
January 4th, 2017, 12:45 PM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,691 Thanks: 1350 
Experience has been my best teacher. You know what I mean by that statement. Some good tips (in the context of regular trig identities, but they still apply to the hyperbolic ones) ... Tips for Trig Identities Basic hyperbolic trig identities you learn, or at the very least become somewhat familiar. These are the ones that get you started ... The Hyperbolic Identities 
January 4th, 2017, 05:46 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,415 Thanks: 1462 
The suggested methods assume you already know that sinh 2x ≡ 2sinh x cosh x and 1 ≡ cosh²x  sinh²x. Have you been taught those identities? If you have, and if x is not zero, coth x = cosh x / sinh x = 2cosh²x / (2sinh x cosh x) $\ \ \ \ \ \ \ \ \ \ $ = (2 + 2sinh²x) / (2sinh x cosh x) $\ \ \ \ \ \ \ \ \ \ $ = 2/sinh 2x + sinh x / cosh x, etc. If you prefer not to assume the identities mentioned above, you can instead use the equations sinh x = (e^x  e^(x))/2 and cosh x = (e^x + e^(x))/2, along with coth x = cosh x / sinh x and tanh x = sinh x / cosh x, etc. 

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