Hyperbolic functions

  • Thread starter Pietair
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  • #1
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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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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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....[itex]1+\tan^2 y=[/itex]___?
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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Hi Pietair! :smile:
If sinhx=tany show x=ln(tany±secy)

The Attempt at a Solution


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

Hint: if sinhx=tany, coshx = … ? :smile:
 
  • #4
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Now, factor out a 4 from inside the square root....[itex]1+\tan^2 y=[/itex]___?
I have no idea...

Should be something like 0.5cos^2y or something I think.
 
  • #5
gabbagabbahey
Homework Helper
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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...
 
  • #6
59
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Thanks a lot, I succeeded by replacing:
[itex]
1+\tan^2 y
[/itex]

for: [itex]
sec^2 y
[/itex]
 

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