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January 1st, 2017, 06:46 AM   #1
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hyperbolic identities

Prove the hyperbolic identities

coth x ≡ cosech 2x + tanh x
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January 1st, 2017, 08:43 AM   #2
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Quote:
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Prove the hyperbolic identities

coth x ≡ cosech 2x + tanh x
check your identity equation. should be ...

$\coth{x} = 2\text{csch}(2x) + \tanh{x}$
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January 1st, 2017, 06:45 PM   #3
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Use the identity $\sinh2x=2\sinh x\cosh x$
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January 4th, 2017, 10:55 AM   #4
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check your identity equation. should be ...

$\coth{x} = 2\text{csch}(2x) + \tanh{x}$
where do i go from there?
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January 4th, 2017, 11:34 AM   #5
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$\coth{x} = 2\text{csch}(2x) + \tanh{x}$

$\coth{x} = \dfrac{2}{\sinh(2x)} + \tanh{x}$

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?
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January 4th, 2017, 11:46 AM   #6
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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.
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January 4th, 2017, 12:45 PM   #7
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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
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January 4th, 2017, 05:46 PM   #8
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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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