show that cosh(t) = (e^t+e^(-t))/2

Question

anonymous · Answer

Given what definition of $\cosh(t)$?

zzr0ck3r · Answer

geomtric using hyperbola

anonymous · Answer

that is?

anonymous · Answer

What are we given about $\cosh(t)$ that we can start out with?

zzr0ck3r · Answer

im not quite sure, she drew a hyperbola and then said the area of a slice would be t/2 where t is the archlength then she said something 1/2cosh(t)sinh(t) - integral from 1 to cosh(t) sqrt(x^2-1) ....

zzr0ck3r · Answer

im not quite sure what was going on...

zzr0ck3r · Answer

one sec im finding my notes

anonymous · Answer

Me neither.  I just know $\large \cosh(t) = \frac{e^x+e^{-x}}{2}$ by definition

zzr0ck3r · Answer

attachment

zzr0ck3r · Answer

these are my notes but they suck because I was lost, this was the first time I ever heard of cosh...

anonymous · Answer

Okay... so if we wanted to find the area, I'd recommend integrating with respect to y, because it is y simple.

zzr0ck3r · Answer

lol im lost

anonymous · Answer

Hmmm, never mind I messed up...

zzr0ck3r · Answer

I have a table that has the integral in it, I just dont understand what Im solving for, a budy of mine told be to sub x for cosh(h) in that integral I posted but I dont get why...

anonymous · Answer

okay can you give me the integral then?

anonymous · Answer

What does $t$ represent?  is it the angle?

zzr0ck3r · Answer

yeah

zzr0ck3r · Answer

yeah one sec
integral of sqrt(u^2-a^2) = u/2sqrt(u^2-a^2) - (a^2/2)ln(a+sqrt(u^2-a^2)) + C

zzr0ck3r · Answer

i think at one point I have to use series expansion of e^x i think...

zzr0ck3r · Answer

thats (u/2)*sqrt(u^2-a^2) - (a^2/2)*ln(a+sqrt(u^2-a^2)) + C

anonymous · Answer

Hmmm, okay so actually if I go back to my integration via y idea:
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anonymous · Answer

To make this a little less complicated:
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