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I've been working on hyperbolic functions lately. All is well, however I seem to come across a couple difficulties here and there when it comes to actually proving hyperbolic identities, thus I'm requesting help from you preferably in the form of hints, and if I still can't possibly understand(hopefully not the case) in the form of an answer.

Prove that $coth^2x-1 \equiv cosech^2x$

I started off by breaking this down into left-hand (LHS) and right-hand (RHS) side, and started off by tackling the LHS.

Double-identities didn't seem to do any trick at this point, I simply knew that $$\coth = \frac{\frac{1}{2}\cosh }{\frac{1}{2}\sinh}$$

And that $$cosech = \frac{1}{\sinh}$$

From then on I replaced the hyperbolic-trig functions with their respective $e$ values and thus plugged it in the original equation, which ends up looking like the following:

$$\frac{\frac{1}{2}(e^x + e^{-x})}{\frac{1}{2}(e^x - e^{-x})} \equiv \frac{1}{\frac{1}{2}(e^x-e^{-x})}$$

From here on I get completely lost, so I've either gone on the completely wrong road, or I'm missing something vital. Either way feel free to correct my errors, I'm quite new to calculus so I'm sorry if I don't follow the correct terms and writing conventions all the time

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3 Answers 3

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$$2\cosh(x)=e^x+e^{-x},2\sinh(x)=e^x-e^{-x}$$

$$\coth^2x-\text{cosech}^2x=\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2-\left(\frac2{e^x-e^{-x}}\right)^2=\frac{(e^x+e^{-x})^2-4}{(e^x-e^{-x})^2}$$

Now, $(e^x+e^{-x})^2-(e^x-e^{-x})^2=4\cdot e^x\cdot e^{-x}=4\iff(e^x+e^{-x})^2-4=(e^x-e^{-x})^2$

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  • $\begingroup$ Thanks alot for this answer, mind if I ask you to elaborate on the bottom part of your answer? Whilst I have a basic idea of what you're doing, it's still fairly foggy $\endgroup$
    – Juxhin
    Sep 27, 2014 at 18:44
  • $\begingroup$ @Juxhin, $$(a+b)^2-(a-b)^2=?$$ $\endgroup$
    – lab bhattacharjee
    Sep 27, 2014 at 18:45
  • $\begingroup$ Not quite sure where you got the $(e^x + e^{-x})^2 - (e^x - e^{-x})^2$ though, I feel like I'm missing something very obvious here $\endgroup$
    – Juxhin
    Sep 27, 2014 at 18:56
  • $\begingroup$ @Juxhin, We need to calculate $$(e^x+e^{-x})^2-4,$$ right? $\endgroup$
    – lab bhattacharjee
    Sep 27, 2014 at 19:02
  • $\begingroup$ The question requires me to prove the identity presented in the question title, therefore need to prove that the left-hand side is equivalent to the right-hand side $\endgroup$
    – Juxhin
    Sep 27, 2014 at 19:06
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$\cot h^2x-1=\operatorname{cosech}^2x$

Proof
$$\coth x=\dfrac {\cosh x}{\sinh x} \implies \coth^2x=\dfrac{\cosh^2x}{\sinh^2x}$$

$$\coth^2x-1=\dfrac {\cosh^2 x}{\sinh^2 x}-1 =\dfrac{\cosh^2x - \sinh^2x}{\sinh^2x}$$

Since $\cosh^2x-\sinh^2x=1$,

$$\coth^2x-1 = \frac{1}{\sinh^2x}=\operatorname{cosech}^2x$$

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the left hand side is equal to $\left(\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right)^2-1=\frac{4}{(e^x+e^{-x})^2}$ the right hand side $\frac{4}{(e^{x}-e^{-x})^2}$

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