1. ## Hyperbolic Function Proof

Prove that $\mathrm{sinh}(i\pi-\theta)=\mathrm{sinh}\theta$.

2. Hi !

Originally Posted by Air
Prove that $\mathrm{sinh}(i\pi-\theta)=\mathrm{sinh}\theta$.

$\sinh x=\frac{e^x-e^{-x}}{2}$

Then, replace x by $i \pi-\theta$, and simplify, using this formula : $e^{i \pi}=\cos \pi+i \sin \pi=-1$

-------------

or :

$\sinh(x)=i \sin(ix)$

---> $\sinh(i \pi-\theta)=i \sin(-\pi-i \theta)=i \sin(-\pi+2 \pi-i \theta)=i \sin(\pi-i\theta)$

But $\sin(\pi-x)=\sin (x)$

So $i \sin(\pi-i \theta)=\dots$

3. Originally Posted by Moo
$e^{i \pi}=\cos \pi+i \sin \pi=-1$
How did you know that the formula equals -1 or is it just a standard answer?

4. Originally Posted by Air
How did you know that the formula equals -1 or is it just a standard answer?
That's because $\cos \pi=-1$ and $\sin \pi=0$

You should know that $e^{ix}=\cos x+i \sin x$

Or just know the formula : $e^{i\pi}+1=0$ which can be found in several signatures in this forum (Chris L T521, kalagota, ... ?)

5. So... I did:

$\frac{e^{i\pi-\theta} - e^{-(i\pi-\theta)}}{2}$

I can simplify $e^{i\pi - \theta}=-1e^{\theta}$ but what about $e^{-(i\pi-\theta)}$?

6. Originally Posted by Air
So... I did:

$\frac{e^{i\pi-\theta} - e^{-(i\pi-\theta)}}{2}$

I can simplify $e^{i\pi - \theta}=-1e^{{\color{red}-}\theta}$ but what about $e^{-(i\pi-\theta)}$?
hehe, that's the way (little mistake in red)

$e^{-(i \pi-\theta)}=e^{-i\pi+\theta}=e^{-i\pi} \cdot e^{\theta}=\frac{1}{e^{i \pi}} \cdot e^{\theta}$

<< I used the rule : $a^{b+c}=a^b \cdot a^c$
And $a^{-b}=\frac{1}{a^b}$

be careful ^^

7. Hello
Originally Posted by Moo
[...]
${\color{red}i}\sinh(x)=\sin(ix)$
[...]
"Little mistake in red."

8. Originally Posted by flyingsquirrel
Hello

"Little mistake in red."

I forgot that