Thread: Help! Deriving tanh(2x) using formulae of sinh(x) and cosh(2x)!!

1. Help! Deriving tanh(2x) using formulae of sinh(x) and cosh(2x)!!

I've been trying this for ages but I just can't do it and i'm guessing its a stupidly simple answer >< please help!!

Using the formulae for sinh(2x) and cosh(2x) in terms of sinh(x) and cosh(x), show that

tanh(2x)= (2tanh(x))/(1+tanh^2(x))

2. i think it is almost similar on how you would derive the non-hyperbolic ones..

3. Hello,

Originally Posted by anthmoo
I've been trying this for ages but I just can't do it and i'm guessing its a stupidly simple answer >< please help!!

Using the formulae for sinh(2x) and cosh(2x) in terms of sinh(x) and cosh(x), show that

tanh(2x)= (2tanh(x))/(1+tanh^2(x))

$\displaystyle \cosh 2x=\cosh^2x+\sinh^2x$
$\displaystyle \sinh 2x=2\cosh x\sinh x$

\displaystyle \begin{aligned} \tanh 2x &=\frac{\sinh 2x}{\cosh 2x} \\ &=\frac{2\cosh x\sinh x}{\cosh^2x+\sinh^2x} \\ &=\frac{2\cosh x\sinh x}{\cosh^2x+\sinh^2x} \cdot \frac{\frac{1}{\cosh^2x}}{\frac{1}{\cosh^2x}} \\ &=\frac{2\cdot \frac{\sinh x}{\cosh x}}{1+\left(\frac{\sinh x}{\cosh x}\right)^2} \end{aligned}

$\displaystyle \implies \boxed{\tanh 2x=\frac{2 \cdot \tanh x}{1+\tanh^2 x}} \blacksquare$

P.S. : I like your nick

4. Haha thankyou!

I like yours too! I had mine first hehe

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tanh(2x)

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