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I have been working on this one for a couple of hours and i just get stuck on every attempt i make.

I have to reduce the formula algebraically:

$\sinh(2 \cdot \sinh^{-1}(y))$

And I just can't seem to do it. I tried using the hyperbolic addition formulas to do something but I just ended up with an even more convoluted expression.

I tried using the addition formula with

$\sinh(x + x) = \cosh(x)\sinh(x) + \sinh(x)+\cosh(x)$

where $x$ is $\sinh^{-1}(y)$,

and then I replaced $\cosh(x)$ and $\sinh(x)$ with their definitions. It did not work.

Can anyone help me out here?

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1 Answer 1

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Your idea of using the identity would actually work.

$\sinh(2x) = 2 \sinh(x) \cosh(x)$ where $x = \sinh^{-1}(y)$

$= 2 y \cosh(x)$.

Now $\cosh(x) = \sqrt{1 + \sinh(x)^2}$.

Can you finish?

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  • $\begingroup$ I have no clue where you are heading sorry, can you explain a bit further? $\endgroup$
    – VictorVH
    Mar 8, 2016 at 0:28
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    $\begingroup$ @VictorVH: Please tell me which line is not clear to you. $\endgroup$
    – user21820
    Mar 8, 2016 at 0:32
  • $\begingroup$ The last two lines. I don't quite understand how you get the results and what to do with them. $\endgroup$
    – VictorVH
    Mar 8, 2016 at 0:35
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    $\begingroup$ @VictorVH: The last identity I stated comes from the basic identity $\cosh(x)^2 - \sinh(x)^2 = 1$. You use it to simplify the result you got earlier using the other identity. $\endgroup$
    – user21820
    Mar 8, 2016 at 0:41
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    $\begingroup$ What's $f(f^{-1}(y))$ for any function $f$ that has an inverse? $\endgroup$
    – Lothar Narins
    Mar 8, 2016 at 2:37

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