Help with project - Freudenstein equation

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Homework Help Overview

The discussion revolves around the Freudenstein equation, as presented in a linked project document. The original poster expresses difficulty in isolating a variable and demonstrating the equation as required in their homework assignment.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss isolating terms in the equation and the algebra involved in summing squares. There are attempts to clarify the steps needed to derive the Freudenstein equation from the given relationships.

Discussion Status

Some participants provide guidance on how to manipulate the equations and apply trigonometric identities. There is an ongoing exploration of the algebraic steps required, with no explicit consensus reached on the final form of the equation.

Contextual Notes

Participants note confusion regarding the algebraic manipulation of trigonometric terms and the proper application of identities. The original poster is required to show their work in a specific format as part of their homework assignment.

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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:[tex]b\sin\beta = c\sin\phi- a\sin\theta[/tex][tex]b\cos\beta = c\cos\phi - a\cos\theta[/tex]The sum of the squares of the left-hand sides is just [itex]b^2(\sin^2\theta + \cos^2\theta) = b^2[/itex]

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 [itex]b^2[/itex] using trig ID's.

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

and seeing how this equation:

[itex]b^2=c^2sin^2\theta+c^2cos^2\theta-a^2sin^2\phi-a^2cos^2\phi[/itex]

gets to this equation:

[itex]R^1cos\theta-R^2cos\phi+R^3-cos(\theta-\phi)[/itex]
 
Last edited:
8point1 said:
Thanks for the reply.

cool, I see how to get [itex]b^2[/itex] using trig ID's.

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

and seeing how this equation:

[itex]b^2=c^2sin^2\theta+c^2cos^2\theta-a^2sin^2\phi-a^2cos^2\phi[/itex]

gets to this equation:

[itex]R^1cos\theta-R^2cos\phi+R^3-cos(\theta-\phi)[/itex]

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

[tex](c\sin\phi- a\sin\theta)^2 = (c\sin\phi- a\sin\theta)(c\sin\phi- a\sin\theta)[/tex]

[tex]= c^2\sin^2\phi -2ac\sin\phi\sin\theta - a^2\sin^2\theta[/tex]
 
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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