1. ## Hyperbolic Function Questions

Hello

How does $(exp[A] - exp[-A])/(exp[A] + exp[-A]) = (exp[2A] - 1)/(exp[2A] + 1)$ ?

How does $sin(jy) = jsinh(y)$ ?

If $-1 < tanh(x) < 1$, why is $ln[tanh(x)] < 0$ ?
Also, why does this mean $tanh(x)$ can not be differentiated? (without using the 'multi-valued' approach and differentiating it).

Thanks

2. #1: $\frac{e^a - e^{-a}}{e^a +e^{-a}} = \frac{e^a - \frac{1}{e^a}}{e^a + \frac{1}{e^a}} \cdot {\color{blue}\frac{e^a}{e^a}} = \cdots$

#2: Use Euler's formula.
\begin{aligned} e^{ix} & = \cos x + i\sin x \\ e^{-ix} & = \cos x - i\sin x \end{aligned} \ \ \overbrace{\Rightarrow}^{x \ = \ iy} \ \sin iy = \frac{1}{2i}\left(e^{-iy} - e^{iy}\right) (Subtracted the second line from the first and divided both sides by i)

Can you carry on from here?

#3: $f(u) = \ln u$ isn't defined for $u < 0$ ... And $f(x) = \tanh x$ is differentiable. In fact, $\frac{d}{dx} \tanh x = \text{sech}^2 x$

3. Thanks.

Regarding #2, I got:

$
e^{ix} - e^{-ix} = 2isin(x)$

$\frac{e^{ix} - e^{-ix}}{2}\ = isin(x)$

Where to from here?

4. Let $x = iy$

Edit: Sorry, I had a typo in my first post. Should've been iy instead of ix using the sub here.

5. You should know $\sin (z) = \frac{1}{{2i}}\left( {e^{iz} - e^{ - iz} } \right)$. That is the standard definition for the complex sine function.
Then it is easy to see: $\sin (iy) = \frac{1}{{2i}}\left( {e^{i\left( {iy} \right)} - e^{ - i\left( {iy} \right)} } \right) = - \frac{i}{2}\left( {e^{ - y} - e^y } \right) = i\left( {\frac{{e^y - e^{ - y} }}{2}} \right) = i\sinh (y)$

6. Thanks, I think I got the result:

$e^{iy} - e^{-iy} = 2isin(y)$

$\frac{e^{iy} - e^{-iy}}{2}\ = isin(y)$

$x = iy$

$\frac{e^{x} - e^{-x}}{2}\ = isin(x/i)$ (1)

$sinh(x) = isin(-ix)$

$\Rightarrow\ isinh(x) = sin(ix)$

But...for (1), how does that equate to $sinh(x)$ if it contains the imaginary part $i$? (I thought only $sin(x)$ dealt with imaginary parts, given the $i$ 's in its definition).

Also, trying to obtain $cosh(x)$:

$\frac{e^{x} + e^{-x}}{2}\ = cos(x/i)$

$\Rightarrow \ cosh(x) = cos(ix)$

Is that correct?

7. No. Try reading over our posts again.

$\sin x = \frac{1}{2i} \left(e^{ix} - e^{-ix}\right)$

Here, we let $x = iy$ to get: $\sin (iy) = \frac{1}{2i}\left(e^{i(iy)} - e^{-i(iy)}\right) = \frac{1}{2i} \left(e^{-y} - e^{y}\right) = -\frac{1}{i} \sinh (y)$

8. Thanks I get it now.