Solve Hyperbolic Functions: Show x=ln(tany±secy)

[Pietair]

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]

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]

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 = … ?

 

[Pietair]

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]

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...

 

[Pietair]

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

for: $sec^2 y$