# Sinh in terms of cosh

1. May 3, 2014

### Jhenrique

Is possible express the sine in terms of cosine and vice-versa:

$\sin(x) = \cos(x-\frac{\pi}{2})$

$\cos(x) = \sin(x+\frac{\pi}{2})$

So, of some way, is possible express the sinh in terms of cosh too and vice-versa?

2. May 3, 2014

### lurflurf

sure

$\sinh(x)=-\imath \, \cosh(x+\pi\,\imath/2) \\ \cosh(x)=-\imath \, \sinh(x+\pi\,\imath/2)$

3. May 4, 2014

### Jhenrique

No exist an identity that no involve imaginary unit?

4. May 4, 2014

### HallsofIvy

Staff Emeritus
No, because that is the whole point of the hyperbolic functions: sinh(x)= i sin(ix) and cosh(x)= cos(ix).

5. May 5, 2014

### Curious3141

Of course, there's the 'obvious' one which you can find by rearranging $\cosh^2 x - \sinh^2 x = 1$, but I don't think that's what you're going for, is it?

6. May 6, 2014

### Jhenrique

no, but thanks

7. May 25, 2014

### lurflurf

Last edited by a moderator: May 6, 2017
8. Jun 10, 2014

### docarlson

Considering the graphs of sinh(x) and cosh(x), they don't have the shift property that sin(x) and cos(x) have for x in set of real numbers.

9. Jun 11, 2014

### mathman

cosh2(x) - sinh2(x) = 1.