Solve The System Of Linears Equations for x and y

[Bashyboy]

Homework Statement

(\cos \theta )x + (\sin \theta )y = 1

and

(-\sin \theta )x + (\cos \theta )y = 0

Homework Equations

The Attempt at a Solution

Evidently the answer is that x = \cos \theta and that y = \sin \theta.

Here is my work:

x = \frac{1 - (\sin \theta )y}{\cos \theta}

Substituting this into the second equation, and simplifying:

y = \frac{\tan \theta}{\sin \theta tan \theta + \cos \theta}

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

 

[LCKurtz]

Homework Statement

(\cos \theta )x + (\sin \theta )y = 1

and

(-\sin \theta )x + (\cos \theta )y = 0

Homework Equations

The Attempt at a Solution

Evidently the answer is that x = \cos \theta and that y = \sin \theta.

Here is my work:

x = \frac{1 - (\sin \theta )y}{\cos \theta}

Substituting this into the second equation, and simplifying:

y = \frac{\tan \theta}{\sin \theta tan \theta + \cos \theta}

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.

 

[HallsofIvy]

Personally, I would not have done the problem that way. Starting from the original equations,
cos(\theta)x+ sin(\theta)y= 1 and
-sin(\theta)x+ cos(\theta)y= 0

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

Then multiply the first equation by sin(\theta) and the second equation by cos(\theta)to get sin(\theta)cos(\theta)x+ sin^2(\theta)y= sin(\theta)
-sin(\theta)cos(\theta)x+ cos^2(\theta)y= 0
Adding gives y= sin(\theta).

 

[Bashyboy]

And so theta will be the parameter to the parametric equations that represent the solution?