# Hyperbolic functions

## Homework Statement

If sinhx=tany show x=ln(tany±secy)

## Homework Equations

sinhx=0.5(e^x-e^(-x))
secy=1/cosy
cosy=0.5(e^y+e^(-y))
tany=(e^(jx)-e^(-jx))/(e^(jx)+e^(-jx))
tany=siny/cosy

## The Attempt at a Solution

0.5e^x -0.5e^-x=tany
0.5e^(2x) -0.5=tany e^x
e^(2x) -2tany e^x -1 = 0
e^(2x) = (2tany±square[(-2tany)^2 + 4] / 2

Is this good so far, and what to do with 4tan^2y?

gabbagabbahey
Homework Helper
Gold Member
e^(2x) -2tany e^x -1 = 0
This equation is quadratic in the variable e^x, not e^(2x), so this:

e^(2x) = (2tany±square[(-2tany)^2 + 4] / 2

Is this good so far, and what to do with 4tan^2y?
should be e^(x) = (2tany±square[(-2tany)^2 + 4] / 2.

Now, factor out a 4 from inside the square root....$1+\tan^2 y=$___?

tiny-tim
Homework Helper
Hi Pietair!
If sinhx=tany show x=ln(tany±secy)

## The Attempt at a Solution

0.5e^x -0.5e^-x=tany…
oooh, so complicated!

Hint: if sinhx=tany, coshx = … ?

Now, factor out a 4 from inside the square root....$1+\tan^2 y=$___?
I have no idea...

Should be something like 0.5cos^2y or something I think.

gabbagabbahey
Homework Helper
Gold Member
I have no idea...

Should be something like 0.5cos^2y or something I think.
Look it up in your table of trig identities. Or better yet, derive it yourself by writing tan^2y in terms of sines and cosines and placing everything over the common denominator...

Thanks a lot, I succeeded by replacing:
$1+\tan^2 y$

for: $sec^2 y$