Eliminating $y$ from the system $cx − sy = 2$ and $sx + cy = 1$, where $c=\cos\theta$, $s=\sin\theta$

Question

We will write $c = \cos\theta$ and $s = \sin\theta$ for ease of notation. Eliminate $y$ from the simultaneous equations $$\begin{align} cx − sy = 2 \\ sx + cy = 1 \end{align}$$

How could you eliminate $y$ from these equations? I have no idea where to start.

Thank you. also how does this prove, it is solvable for all values of sin and cos.

Answer 1

Multiply the first equation by $c$ to obtain: $c^2x-scy=2c$ and the second equation by $s$ to get: $s^2x+scy=s$. Now add the 2 equations term by term to obtain $(c^2+s^2)x=2c+s$. Using the trigonometric identity $\sin(x)^2+\cos(x)^2=1$ and you see that $x=2c+s$. Use then one of the original equations, substitute your value for $x$ and see that, after simplification $y=c-2s$.

$x$ and $y$ are defined $\forall \theta$.

Hope this helps.

Answer 2

After eliminating $y$ usual way ( mentioned by A. Goodier in comments ) you get $$ 2 \cos \theta + \sin \theta =x $$

To find range for $x:$ differentiate, set to zero and solve

$$ -2 \sin \theta+ \cos \theta =0$$

$\theta=\tan^{-1} \frac12;$ plug this into the expression to evaluate max/min range values, it comes to $ \pm \sqrt 5.$

$x$ must be in the interval $ (\sqrt 5 > x >-\sqrt 5) $ to have a real $ \theta$ angle solution ( that would be periodic).