Solve The System Of Linears Equations for x and y

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Homework Help Overview

The discussion revolves around solving a system of linear equations involving trigonometric functions, specifically equations that relate \(x\) and \(y\) to \(\theta\). The equations presented are \((\cos \theta)x + (\sin \theta)y = 1\) and \((- \sin \theta)x + (\cos \theta)y = 0\).

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods for solving the equations, including substitution and manipulation of the original equations. Some express confusion about the complexity of their attempts, while others suggest alternative approaches, such as using determinants.

Discussion Status

The discussion is active, with participants sharing their methods and reasoning. Some guidance has been offered regarding simplification techniques and alternative strategies, but there is no explicit consensus on the best approach yet.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the methods they can use or the assistance they can seek. There is also a mention of the parameterization of the solution in terms of \(\theta\), indicating a potential exploration of parametric equations.

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Homework Statement


[itex](\cos \theta )x + (\sin \theta )y = 1[/itex]

and

[itex](-\sin \theta )x + (\cos \theta )y = 0[/itex]

Homework Equations





The Attempt at a Solution



Evidently the answer is that [itex]x = \cos \theta[/itex] and that [itex]y = \sin \theta[/itex].

Here is my work:

[itex]x = \frac{1 - (\sin \theta )y}{\cos \theta}[/itex]

Substituting this into the second equation, and simplifying:

[itex]y = \frac{\tan \theta}{\sin \theta tan \theta + \cos \theta}[/itex]

I then took this equation and back-substituted into [itex]x = \frac{1 - (\sin \theta )y}{\cos \theta}[/itex], hoping that everything would simplify such that [itex]x= \cos \theta[/itex]; however, things began to look quite messy. How am I to solve this problem?
 
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Bashyboy said:

Homework Statement


[itex](\cos \theta )x + (\sin \theta )y = 1[/itex]

and

[itex](-\sin \theta )x + (\cos \theta )y = 0[/itex]

Homework Equations





The Attempt at a Solution



Evidently the answer is that [itex]x = \cos \theta[/itex] and that [itex]y = \sin \theta[/itex].

Here is my work:

[itex]x = \frac{1 - (\sin \theta )y}{\cos \theta}[/itex]

Substituting this into the second equation, and simplifying:

[itex]y = \frac{\tan \theta}{\sin \theta tan \theta + \cos \theta}[/itex]

Multiply the numerator and denominator of that last equation by ##\cos\theta## and you will have it. Much easier to use determinants in the first place though.
 
Personally, I would not have done the problem that way. Starting from the original equations,
[itex]cos(\theta)x+ sin(\theta)y= 1[/itex] and
[itex]-sin(\theta)x+ cos(\theta)y= 0[/itex]

Multiply the first equation by [itex]cos(theta)[/itex] and the second equation by [itex]-sin(\theta)[/itex] to get
[itex]cos^2(\theta)x+ sin(\theta)cos(\theta)y= cos(\theta)[/itex]
[itex]sin^2(\theta)x- sin(\theta)cos(\theta)y= 0[/itex]
and then add:
[itex]x= cos(\theta)[/itex]

Then multiply the first equation by [itex]sin(\theta)[/itex] and the second equation by [itex]cos(\theta)[/itex]to get [itex]sin(\theta)cos(\theta)x+ sin^2(\theta)y= sin(\theta)[/itex]
[itex]-sin(\theta)cos(\theta)x+ cos^2(\theta)y= 0[/itex]
Adding gives [itex]y= sin(\theta)[/itex].
 
And so theta will be the parameter to the parametric equations that represent the solution?
 

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