# Solve The System Of Linears Equations for x and y

1. Jun 27, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
$(\cos \theta )x + (\sin \theta )y = 1$

and

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

2. Relevant equations

3. 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?

2. Jun 27, 2013

### LCKurtz

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.

3. Jun 27, 2013

### HallsofIvy

Staff Emeritus
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
$cos^2(\theta)x+ sin(\theta)cos(\theta)y= cos(\theta)$
$sin^2(\theta)x- sin(\theta)cos(\theta)y= 0$
$x= cos(\theta)$

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 $y= sin(\theta)$.

4. Jun 27, 2013

### Bashyboy

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