# P4 Identities (er... Hyperbolas!)

Well, I think they are called 'Indentities', but I am not 100% sure.

My teacher gave us this to solve. But I keep getting stuck.

Prove,

tanh 2x = (2 tanh x) / (1 + tan h^2 x)
With the RHS side I have got as far as,

tanh 2x = (2 * (e^x - e^-x/e^x + e^-x)) / (1 + (e^2x - 2 + e^-2x/e^2x + 2 + e^-2x))

tanh 2x = (2 * (e^x - e^-x/e^x + e^-x)) / ((e^2x - 2 + e^-2x + e^2x + 2 + e^-2x/e^2x + 2 + e^-2x))

tanh 2x = (2 * (e^x - e^-x/e^x + e^-x)) / ((2e^2x + 2e^-2x /e^2x + 2 + e^-2x))

tanh 2x = (2 * (e^x - e^-x/e^x + e^-x)) / (( 2 (e^x + e^-x) /e^2x + 2 + e^-2x))

Thanks
It would be better to avoid using the definitions in terms of $e^x, e^{-x}$ as with squares and multiples of angles it can get rather messy.
Have you tried expanding out $\tanh 2x$ by using $\sinh 2x =2\sinh x \cosh x$ and $\cosh 2x = \cosh ^2 x + \sinh ^2 x$?
erm yea, i couldn't get anywhere with it though.
Try writing it out on a bit of scrap paper.
What does $\frac{\sinh 2x}{\cosh ^2 x}$ equal in terms of single values of x? Does this help at all?
Im not sure lol. We have just been told to do it using the e's
I was going to suggest writing $\tanh 2x = \frac{\sinh 2x}{\cosh 2x} = \frac{2\sinh x \cosh x}{\cosh ^2 x + \sinh ^2 x} = \frac{2\tanh x}{1+\tanh ^2 x}$
You can do it using $e^x$'s if you really want to but it won't like very nice.
Vazzyb
tanh 2x = (2 tanhx) / (1 + tanh^2x)

tanh2x = (sinh2x)/(cosh2x)
sinh2x = 2sinhxcoshx
cosh2x = cosh^2x + sinh^2x = 2sinh^2x + 1

:. tanh2x = (sinh2x)/(cosh2x)
= (2sinhxcoshx)/(1 + 2sinh^2x)
= (2sinhxcoshx)/[1 + 2tan^2x.cosh^2x]
= (2sinhxcoshx)/[cosh^2x - sinh^2x + 2tanh^2x.cosh^2x]
= (2sinhxcoshx)/[cosh^2x(1 - tanh^2x + 2tanh^2x)]
= (2sinhxcoshx)/[cosh^2x(1 + tanh^2x)]
= (2sinhx)/[coshx(1 + tanh^2x)]
= (2tanhx)/(1 + tanh^2x)
Gaz031
I was going to suggest writing $\tanh 2x = \frac{\sinh 2x}{\cosh 2x} = \frac{2\sinh x \cosh x}{\cosh ^2 x + \sinh ^2 x} = \frac{2\tanh x}{1+\tanh ^2 x}$
You can do it using $e^x$'s if you really want to but it won't like very nice.

I don't think you can skip so much regarding the last step tbh, it takes quite a bit of simplifying etc. to get it down to the identity (see mine for the extra work needed)
It requires barely any simplification - you simply divide top and bottom of the fraction by $\cosh ^2 x$ and the result follows.
Vazzyb
Well, I think they are called 'Indentities', but I am not 100% sure.

My teacher gave us this to solve. But I keep getting stuck.

Prove,

tanh 2x = (2 tanh x) / (1 + tan h^2 x)
With the RHS side I have got as far as,

tanh x= (e^x-e^-x)/(e^x+e^-x)
tanh^2x=(e^2x-2+e^-2x)/(e^2x+2+e^-2x)
1+tanh^2(x)=(e^2x+2+e^-2x+e^2x-2+e^-2x)/(e^x+e^-x)^2
(2 tanh x) / (1 + tan h^2 x)
= 2(e^x-e^-x)/(e^x+e^-x)/2(e^2x+e^-2x)/(e^x+e^-x)^2
=(e^x-e^-x)(e^x+e^-x)/(e^2x+e^-2x)
=e^2x-e^-2x/e^2x+e^-2x
=tanh 2x
Right, I get it. From the e-version though :P

Thanks very much,

Varun

3rd week of Further Maths -
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