# Math Help - cosh

1. ## cosh

can someone please explain to me how they got from

$\frac{-1}{4i}(e^{-1}+e^1)$

to

$\frac{i}{2}cosh(1)$

coz if i did it i got

$-\frac{1}{2i} cosh (1)$

soo sorry im so lost

2. Originally Posted by Dgphru
can someone please explain to me how they got from

$\frac{-1}{4i}(e^{-1}+e^1)$

to

$\frac{i}{2}cosh(1)$

coz if i did it i got

$-\frac{1}{2i} cosh (1)$

soo sorry im so lost
It's the same thing!

$-\frac{1}{2i} = -\frac{i}{2i^2} = - \frac{i}{2\cdot(-1)} = \frac{i}{2}$

3. hahah i can be so stupid ><