Solving a pair of equations across a data set

Question

I have the following pair of equations:

$a = x\cos(k) - y\sin(k),\ b = x\sin(k) + y\cos(k)$

I know that all variables but $k$ are in the set $\mathbb{N}^0$.

Given the above, if I know the value of $a$, $b$, and $k$, how do I compute $x$ and $y$?

Answer 1

If we know $a$, $b$, and $k$, we have the matrix equations $$ \begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}\cos(k)&-\sin(k)\\\sin(k)&\cos(k)\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\ \Downarrow\\ \begin{bmatrix}\cos(k)&\sin(k)\\-\sin(k)&\cos(k)\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix} $$ which solve for $x$ and $y$.

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For Those Who Haven't Seen Matrices Before

The first matrix equation above is a restatement of the question. The second matrix equation is the first multiplied by the inverse of the $2\times2$ matrix, and represents the solution $$ \begin{align} x&=\hphantom{-}a\cos(k)+b\sin(k)\\ y&=-a\sin(k)+b\cos(k) \end{align} $$ If you plug in the values for $a$ and $b$ from the question, you will see that $$ \begin{align} \hphantom{-}a\cos(k)+b\sin(k)&=\hphantom{-}\overbrace{(x\cos(k)-y\sin(k))}^a\cos(k)+\overbrace{(x\sin(k)+y\cos(k))}^b\sin(k)=x\\ -a\sin(k)+b\cos(k)&=-(x\cos(k)-y\sin(k))\sin(k)+(x\sin(k)+y\cos(k))\cos(k)=y \end{align} $$ since $\cos^2(k)+\sin^2(k)=1$.