# Throw whatever you have against these 2 equations

1. Nov 23, 2008

### andrewm

This isn't my homework: I'm doing some physics research and I'm stuck at a simple 2 equations. I want to solve these equations

$$A \cos(\gamma) \sinh(\theta) = \lambda - B \cosh(\theta)$$
$$A \cos(\gamma) \cosh(\theta) = A \sin(\gamma) - B \sinh(\theta)$$

I'd like to know if there's any way I can find $$\lambda$$ if I start with A and B known. I'd be happy to do this numerically, but I can't see how I would. I've tried monkeying with the algebra for a while.

All that my undergrad math tells me is that there should be a solution since there are 2 equations, 2 unknowns.

Is there any way I can solve this numerically, or approximate the solution by hand, or even show that there is a solution?

I'm stumped!

2. Nov 23, 2008

### 03myersd

There are 4 unknowns in this. A, B, Gamma, Theta.

There may be a way of solving this but not with any form of simulataneous equations of matrices i wouldn't think. Don't hold me to that though.

3. Nov 23, 2008

### andrewm

A and B are known parameters. Indeed, I have already tabulated them numerically.

4. Nov 23, 2008

### 03myersd

Sorry that was me misreading the post. Another problem is still there. Even though we have two equations we have three variables. (I missed out lambda last time). You would need 3 equations for this to work.

5. Nov 23, 2008

### andrewm

Of course, thanks. I'll try to find a new constraint.

6. Nov 23, 2008

### arildno

At the very least, you may square&subtract:
$$A^{2}\cos^{2}(\gamma) \cosh^{2}(\theta)-A^{2} \cos^{2}(\gamma) \sinh^{2}(\theta)= (A \sin(\gamma) - B \sinh(\theta))^{2}-(\lambda - B \cosh(\theta) )^{2}$$
Which can be simplified to:
$$A^{2}\cos(2\gamma)=2B\lambda\cosh(\theta)-2AB\sin(\gamma)\sinh(\theta)-(B^{2}+\lambda^{2})$$
Maybe this equation can be used for something, or maybe not.

7. Nov 23, 2008

### HallsofIvy

You still have three unknowns, $\lambda$, $\theta$, and $\gamma$ with only two equations.

8. Nov 24, 2008

### Office_Shredder

Staff Emeritus
If you're just looking for any solution...

To start, pick gamma=pi/2. Then your equations reduce to

$$\lambda -Bcosh( \theta ) = 0$$
$$A - Bsinh (\theta ) = 0$$

Based on the second equation, you find theta, then use the first equation to find lambda.

9. Nov 24, 2008

### Gerenuk

I'd bring all thetas on one side and then square and subtract. The result is
$$\cos 2\gamma=\frac{B^2-\lambda^2}{A^2}$$
but please check it for yourself.