Prove that $\displaystyle \mathrm{sinh}(i\pi-\theta)=\mathrm{sinh}\theta$.
Thanks in advance.
Hi !
$\displaystyle \sinh x=\frac{e^x-e^{-x}}{2}$
Then, replace x by $\displaystyle i \pi-\theta$, and simplify, using this formula : $\displaystyle e^{i \pi}=\cos \pi+i \sin \pi=-1$
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or :
$\displaystyle \sinh(x)=i \sin(ix)$
---> $\displaystyle \sinh(i \pi-\theta)=i \sin(-\pi-i \theta)=i \sin(-\pi+2 \pi-i \theta)=i \sin(\pi-i\theta)$
But $\displaystyle \sin(\pi-x)=\sin (x)$
So $\displaystyle i \sin(\pi-i \theta)=\dots$
That's because $\displaystyle \cos \pi=-1$ and $\displaystyle \sin \pi=0$
You should know that $\displaystyle e^{ix}=\cos x+i \sin x$
Or just know the formula : $\displaystyle e^{i\pi}+1=0$ which can be found in several signatures in this forum (Chris L T521, kalagota, ... ?)
hehe, that's the way (little mistake in red)
$\displaystyle e^{-(i \pi-\theta)}=e^{-i\pi+\theta}=e^{-i\pi} \cdot e^{\theta}=\frac{1}{e^{i \pi}} \cdot e^{\theta}$
<< I used the rule : $\displaystyle a^{b+c}=a^b \cdot a^c$
And $\displaystyle a^{-b}=\frac{1}{a^b}$
be careful ^^