Let $\mathbf{D}$ be a matrix representing a dilation with scale factor $k > 0,$ and let $\mathbf{R}$ be a matrix representing a rotation about the origin by an angle of $\theta$ counter-clockwise.  If
\[\mathbf{R} \mathbf{D} = \begin{pmatrix} 8 & -4 \\ 4 & 8 \end{pmatrix},\]then find $\tan \theta.$
We have that $\mathbf{D} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ and $\mathbf{R} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix},$ so
\[\mathbf{R} \mathbf{D} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = \begin{pmatrix} k \cos \theta & -k \sin \theta \\ k \sin \theta & k \cos \theta \end{pmatrix}.\]Thus, $k \cos \theta = 8$ and $k \sin \theta = 4.$  Dividing these equations, we find $\tan \theta = \boxed{\frac{1}{2}}.$
\frac{1}{2}
Level 3
Precalculus
Problem: Let $\mathbf{D}$ be a matrix representing a dilation with scale factor $k > 0,$ and let $\mathbf{R}$ be a matrix representing a rotation about the origin by an angle of $\theta$ counter-clockwise.  If
\[\mathbf{R} \mathbf{D} = \begin{pmatrix} 8 & -4 \\ 4 & 8 \end{pmatrix},\]then find $\tan \theta.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\tan \theta$, we first need to understand the matrices involved. The dilation matrix $\mathbf{D}$ can be expressed as $\mathbf{D} = k \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$. The rotation matrix $\mathbf{R}$ is given by $\mathbf{R} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$. 

The product of these two matrices is:
\[
\mathbf{R} \mathbf{D} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = \begin{pmatrix} k \cos \theta & -k \sin \theta \\ k \sin \theta & k \cos \theta \end{pmatrix}.
\]

We know that this product equals the matrix $\begin{pmatrix} 8 & -4 \\ 4 & 8 \end{pmatrix}$. Therefore, we can set up the following equations based on the corresponding entries of the matrices:
1. \( k \cos \theta = 8 \)
2. \( -k \sin \theta = -4 \) (which simplifies to \( k \sin \theta = 4 \))

From these equations, we can express $k$ in terms of $\cos \theta$ and $\sin \theta$:
- From the first equation, we have \( k = \frac{8}{\cos \theta} \).
- Substituting this into the second equation gives:
\[
\frac{8}{\cos \theta} \sin \theta = 4.
\]
This simplifies to:
\[
8 \sin \theta = 4 \cos \theta.
\]
Dividing both sides by 4 yields:
\[
2 \sin \theta = \cos \theta.
\]
Rearranging gives:
\[
\frac{\sin \theta}{\cos \theta} = \frac{1}{2}.
\]
Thus, we find that:
\[
\tan \theta = \frac{1}{2}.
\]

The final answer is $\frac{1}{2}$.<|END_OF_LATENT|> The final answer is $\frac{1}{2}$.