cosh

**Dgphru** · May 19th 2010, 11:23 PM

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

**drumist** · May 19th 2010, 11:26 PM

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!

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

**Dgphru** · May 19th 2010, 11:32 PM

hahah i can be so stupid ><

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