Help with project - Freudenstein equation

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The discussion revolves around solving the Freudenstein equation, specifically isolating terms in β and demonstrating the equation's derivation. The user is struggling with the algebra involved in summing the squares of the equations after isolating β. Participants clarify that using trigonometric identities is essential for simplifying the equations and correctly applying the FOIL method for squaring terms. The conversation emphasizes the importance of careful algebraic manipulation and the correct application of trigonometric identities to reach the desired equation. Overall, the thread provides collaborative assistance in navigating the complexities of the problem.
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Homework Statement


Its easier to see if I link the problem:
http://faculty.olympic.edu/jjbrown/documents/ENGR%20111/ENGR%20111%20Project%201%202011.pdf

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
 
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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\thetab\cos\beta = c\cos\phi - a\cos\thetaThe 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.
 
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:
8point1 said:
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)

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

Should that last term be +?
 
8point1 said:
You're right, I didn't foil the right side. gets confusing with all the sines and cosines.

Should that last term be +?

Yes.
 
Okay, thanks for your help!
 
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