Sin x =cosh x

1. Aug 12, 2011

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.

2. Aug 12, 2011

SteamKing

Staff Emeritus
Express sin and cosh in their exponential form.

3. Aug 12, 2011

Fredrik

Staff Emeritus
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.

\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.

Last edited: Aug 12, 2011
4. Aug 13, 2011

dimension10

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.

5. Aug 13, 2011

I like Serena

I suggest substituting y=ex...

6. Aug 13, 2011

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?

7. Aug 13, 2011

gb7nash

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.

8. Aug 13, 2011

D H

Staff Emeritus
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:

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

9. Aug 13, 2011

D H

Staff Emeritus
I suggest setting $x=u+iv$.

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

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

Last edited: Aug 13, 2011
10. Aug 13, 2011

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.

Last edited by a moderator: Aug 13, 2011
11. Aug 13, 2011

Fredrik

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

Last edited: Aug 13, 2011
12. Aug 13, 2011

ForMyThunder

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

13. Aug 13, 2011

dimension10

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

But then x would pi/6 which is false.

Last edited: Aug 13, 2011