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I have to prove the following: $$[\sinh(a\tau'')-\sinh(a\tau')-ia\epsilon]^2-[\cosh(a\tau'')-\cosh(a\tau')]^2=4[\sinh(\frac{a}{2}(\tau''\tau')-i\epsilon)]^2$$

we know that:

$a$ is a real positive number

$\epsilon\rightarrow0$, so $\cosh(i\epsilon)=1$, and $\sinh(i\epsilon)=i\epsilon$

I have tried using the trigonometric formulas:

$$\sinh{(A)}-\sinh{(B)}=2\sinh{\left(\frac{A-B}{2}\right)}\cosh{\left(\frac{A+B}{2}\right)}$$

$$\cosh{(A)}-\cosh{(B)}=2\sinh{\left(\frac{A+B}{2}\right)}\sinh{\left(\frac{A-B}{2}\right)}$$

but I couldn't prove it. Could you please help me by giving the solution or hints? Thank you for your time.

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Quite simple: start from the addition formulae: $$\begin{cases}\sinh(a+b)=\sinh a\cosh b+\sinh b\cosh a\\ \sinh(a-b)=\sinh a\cosh b-\sinh b\cosh \end{cases}$$ whence $\,2\sinh b\cosh a=\sinh(a+b)-\sinh(a-b)$.

Now set $A=a+b$, $\,B=a-b$, which is equivalent to $\,a=\dfrac{a+b}2,\enspace b=\dfrac{A-B}2$, and you get the first factorisation formula.

Smae method for the second formula, starting from the addition formulae for $\cosh$.

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  • $\begingroup$ Sorry, perhaps I did not explain it well. I need to prove the first equality, not the trigonometric formulae. $\endgroup$ – Fischer Jun 11 '15 at 13:17

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