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

In summary, to solve for x when sinhx=tany, you can use the equation x=ln(tany±secy) and factor out a 4 from inside the square root of the quadratic equation. Then, use the trigonometric identity sec^2 y=1+tan^2 y to simplify the equation and solve for x.
  • #1
Pietair
59
0

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?
 
Physics news on Phys.org
  • #2
Pietair said:
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
Hi Pietair! :smile:
Pietair said:
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
gabbagabbahey said:
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
Pietair said:
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
Thanks a lot, I succeeded by replacing:
[itex]
1+\tan^2 y
[/itex]

for: [itex]
sec^2 y
[/itex]
 

What are hyperbolic functions?

Hyperbolic functions are mathematical functions that are closely related to trigonometric functions. They are defined in terms of the exponential function and involve the use of the hyperbolic sine, cosine, tangent, cotangent, secant, and cosecant.

What is the equation to solve for x in "ln(tany±secy)"?

The equation to solve for x in "ln(tany±secy)" is x=ln(tany±secy). This equation involves the natural logarithm of a trigonometric function, tangent, and a hyperbolic function, secant.

How do I solve for x in "ln(tany±secy)"?

To solve for x in "ln(tany±secy)", you can use the properties of logarithms and hyperbolic functions to simplify the expression. Then, you can use algebraic techniques to isolate x on one side of the equation.

Are there any restrictions on the values of y for this equation?

Yes, there are restrictions on the values of y for this equation. Since the natural logarithm function and the tangent function are only defined for positive values, the values of y must be greater than 0. Additionally, the secant function is not defined for certain values, such as 0 and π/2, so the values of y cannot be those values either.

What is the significance of solving "ln(tany±secy)"?

Solving "ln(tany±secy)" can be used in various applications, such as in physics, engineering, and economics. It can help in finding the solutions to certain problems involving exponential growth or decay, as well as in finding the optimal solutions for certain mathematical models.

Similar threads

  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
704
  • Calculus and Beyond Homework Help
Replies
1
Views
781
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
708
  • Calculus and Beyond Homework Help
Replies
2
Views
794
  • Calculus and Beyond Homework Help
Replies
1
Views
663
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top