Solve The System Of Linears Equations for x and y

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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?