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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))

Please help,

Thanks

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))

Please help,

Thanks

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.

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)

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,

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

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- A level mathematics
- IAL Edexcel- Maths- do grades usually drop in A2
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- what to do ????
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- Unit 7 p4
- Can I bring my B upto an A in chemistry 9701?
- Retaking Exam

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