## Elementary Linear Algebra 7th Edition

The solution of this system is unique and it is: $$x=\cos\theta-\sin\theta,\quad y= \cos\theta + \sin\theta.$$
Here we can treat sines and cosines as ordinary coefficients and we will solve this system as we would solve any system of two equations with two unknowns. Step 1: Multiply the first equation by $\sin\theta$ and the second equation by $\cos\theta$: \begin{align*} (\sin\theta\cos\theta)x+(\sin^2\theta)y=&\sin\theta\\ (-\sin\theta\cos\theta)x+(\cos^2\theta)y =& \cos\theta. \end{align*} Step 2: Add these two equations together to eliminate $x$ and find $y$: $$(\sin\theta\cos\theta)x +(-\sin\theta\cos\theta)x + (\sin^2\theta)y +(\cos^2\theta)y = \sin\theta+\cos\theta$$ which becomes $$y(\sin^2\theta+\cos^2\theta) = \sin\theta+\cos\theta.$$ Remember that for any $\theta$ we have that $\sin^2\theta+\cos^2\theta = 1$ so this simplifies to $$y=\sin\theta + \cos\theta$$ Step 3: Now let us do the similar thing as in step 1. Multiply the first equation by $\cos\theta$ and the second one by $\sin\theta$: \begin{align*} (\cos^2\theta)x+ (\sin\theta\cos\theta) y = &\cos\theta\\ (-\sin^2\theta)x+(\sin\theta\cos\theta)y = &\sin\theta \end{align*} Step 4: Now subtract these equations to eliminate $y$ and find $x$ : $$(\cos^2\theta)x-(-\sin^2\theta)x + (\sin\theta\cos\theta) y -(\sin\theta\cos\theta)y =\cos\theta- \sin\theta$$ This then gives $$x(\sin^2\theta+\cos^2\theta) = \cos\theta-\sin\theta.$$ Using the fact that $\sin^2\theta +\cos^2\theta = 1$ we again get simpler expression: $$x=\cos\theta-\sin\theta.$$