# Help with project - Freudenstein equation

1. Feb 3, 2012

### 8point1

1. The problem statement, all variables and given/known data
Its easier to see if I link the problem:
http://faculty.olympic.edu/jjbrown/documents/ENGR%20111/ENGR%20111%20Project%201%202011.pdf

3. The attempt at a solution

I don't have a problem finding the answers, but the other steps we need to show are giving me a hard time. My hang up is the sentence: "By isolating in terms of β, then adding the squares of those equations one obtains the Freudenstein equation (you are to show this)."

I tried working the above equation to solve for β, but I'm just not seeing it. Any help would be greatly appreciated. Thanks

2. Feb 3, 2012

### cepheid

Staff Emeritus
Just follow the steps that were given. Isolating the terms in beta means moving them to one side and everything else to the other side. Hence the equations become:$$b\sin\beta = c\sin\phi- a\sin\theta$$$$b\cos\beta = c\cos\phi - a\cos\theta$$The sum of the squares of the left-hand sides is just $b^2(\sin^2\theta + \cos^2\theta) = b^2$

So, the sum of the squares of the right-hand sides has to be equal to that. Some trig identities are definitely going to come into play here.

3. Feb 3, 2012

### 8point1

Thanks for the reply.

cool, I see how to get $b^2$ using trig ID's.

But adding the sums of the right hand I'm not seeing the algebra.

and seeing how this equation:

$b^2=c^2sin^2\theta+c^2cos^2\theta-a^2sin^2\phi-a^2cos^2\phi$

gets to this equation:

$R^1cos\theta-R^2cos\phi+R^3-cos(\theta-\phi)$

Last edited: Feb 3, 2012
4. Feb 3, 2012

### cepheid

Staff Emeritus
First of all, I don't think you're really squaring the right-hand sides correctly:

$$(c\sin\phi- a\sin\theta)^2 = (c\sin\phi- a\sin\theta)(c\sin\phi- a\sin\theta)$$

$$= c^2\sin^2\phi -2ac\sin\phi\sin\theta - a^2\sin^2\theta$$

5. Feb 3, 2012

### 8point1

You're right, I didn't foil the right side. gets confusing with all the sines and cosines.

Should that last term be +?

6. Feb 3, 2012

### cepheid

Staff Emeritus
Yes.

7. Feb 3, 2012

### 8point1

Okay, thanks for your help!