How can we find the true solution for cosh x = tanh x?

[dimension10]

According to Wolfram Alpha, if sin x =cosh x then x=-(\frac{3\pi}{4}+\frac{3i\pi}{4})

But how do you derive this?

Also, what is the solution to cos x = sinh x in exact form?


Thanks.

 

[SteamKing]

Express sin and cosh in their exponential form.

 

[Fredrik]

According to Wolfram Alpha, if sin x =cosh x then x=-(\frac{3\pi}{4}+\frac{3i\pi}{4})

But how do you derive this?

Also, what is the solution to cos x = sinh x in exact form? Thanks.

This is not true. What's true is that if x=-(\frac{3\pi}{4}+\frac{3i\pi}{4}), then sin x=cosh x. Note that "if A, then B" (where A and B are statements, not numbers) is not equivalent to "if B, then A". For example, "if x=dimension10, then x is human" is not equivalent to "if x is human, then x=dimension10". "if A, then B" is however equivalent to "if not B, then not A". This fact is very useful in proofs.

I would start with the formulas \begin{align}
\sin x=\frac{e^{ix}-e^{-ix}}{2i}\\
\cosh x=\frac{e^{x}-e^{-x}}{2}
\end{align} This is what SteamKing had in mind. I'm including them in this post because I suspect that you don't know them. I realize that you're going to need more than this, but I don't have time to think about this now.

If you type "solve sin x=cosh x" into Wolfram Alpha, it will tell you all the solutions in the complex plane.

 

[dimension10]

I would start with the formulas

\sin x=\frac{e^{ix}-e^{-ix}}{2i}

\cosh x=\frac{e^{x}-e^{-x}}{2}

This is what SteamKing had in mind. I'm including them in this post because I suspect that you don't know them.

I do know the exponential forms. However, they do not seem to be helpful in finding x. I even used Euler's formula, but nothing works.

 

[I like Serena]

I do know the exponential forms. However, they do not seem to be helpful in finding x. I even used Euler's formula, but nothing works.

I suggest substituting y=e^x...

 

[HallsofIvy]

But that is still going to give you a messy equation. With the substitution y= e^x, I get y^{1+ i}- y^{1- i}= iy.

What next?

 

[gb7nash]

But that is still going to give you a messy equation. With the substitution y= e^x, I get y^{1+ i}- y^{1- i}= iy.

What next?

This might work:

Dividing out y, we get:

y^{i}- y^{-i}= i. (since y != 0)

Setting z = y^{i}, you'll get z - \frac{1}{z} = i, so:

z^2 - 1 = iz and z^2 - iz - 1

From here, you should get:

z = \frac{i \pm \sqrt{-1 + 4}}{2}

Now, we managed to solve for y^i, but this is where I get stuck.

 

[D H]

If you type "solve sin x=cosh x" into Wolfram Alpha, it will tell you all the solutions in the complex plane.

Yes, but it does so in a way that shows lack of insight (Mathematica's results ofttimes are messy; but then again, so are Maple's and Maxima's) . Mathematica doesn't recognize that there is a unifying way to represent these solutions:

x = (n+1/4)(1\pm i)\pi,\quad n\in\mathbb{Z}

 

[D H]

Express sin and cosh in their exponential form.

I suggest substituting y=e^x...

I suggest setting x=u+iv.

After just a bit of work you should arrive at
\begin{align}<br /> \sin x &amp;= \sin u \cosh v + i \cos u \sinh v \\<br /> \cosh x &amp;= \cos v \cosh u + i \sin v \sinh u<br /> \end{align}
Equating the real and imaginary parts yields
\begin{align}<br /> \sin u \cosh v &amp;= \cos v \cosh u \\<br /> \cos u \sinh v &amp;= \sin v \sinh u<br /> \end{align}

And that is more than enough of a hint for now.

 

[uart]

Another way to do it is to recognize that cosh(x) = cos(ix)

Then you have sin(x) = cos(ix) and from there you can use the complementary relationship between sin and cos.

 

[Fredrik]

Last edited by D H; T at 03:38 PM.. Reason: Too much help. We don't know if this is homework.

It's not homework. dimension10 is too young to get this sort of problems in school.

 

[ForMyThunder]

Another way to do it is to recognize that cosh(x) = cos(ix)

Then you have sin(x) = cos(ix) and from there you can use the complementary relationship between sin and cos.

That's a nice way. Probably wouldn't get any easier.

 

[dimension10]

This might work:

Dividing out y, we get:

y^{i}- y^{-i}= i. (since y != 0)

Setting z = y^{i}, you'll get z - \frac{1}{z} = i, so:

z^2 - 1 = iz and z^2 - iz - 1

From here, you should get:

z = \frac{i \pm \sqrt{-1 + 4}}{2}

Now, we managed to solve for y^i, but this is where I get stuck.

Thanks. Strangely, i \arcsin z is the solution for cosh x = tanh x.

But then x would pi/6 which is false.