Adding $\cos \theta$ and $\sin \theta$ of Perpendicular Vectors In order to derive 3D rotation formula known as Rodriguez formula, I am trying to figure out a vector calculation given in the book by Szeliski pg. 38, where for the figure given below

The context here is rotation: $u$ is the rotated form of $v$ around the axis $\hat{n}$ after a rotation of $\theta$ degrees. The magnitude of $u$ and $v$ would be the same. $\hat{n}$ is a unit vector and is normal to the plane shown above. 
After some derivations for $v_\perp$ and $v_x$ (which I understand) the author uses 
$$u_\perp = \cos \theta v_\perp + \sin \theta v_x$$
where $v_x$ is simply $v_\perp$ rotated 90 degrees. The formula above is the part I do not understand - I guess I missing an identity for triangles. 
Any ideas?
Thanks,
 A: It looks like $v_x$ has been obtained from $v_\perp$ by rotation through $\pi/2$ in the direction from $v_1$ toward $u_\perp.$ If (as it is here) the vectors $v_\perp, v_x$ have the same length and are orthogonal, then polar coordinates may be set up in their plane by the usual formula $r(v_1 \cos t, v_\perp \sin t)$ where $t$ (the polar angle) is being measured from $v_1$ towards $v_\perp.$
Here $r$ is just a scale factor to enable representing all points (except the origin) in that polar form. For your diagram, you can take $r=1$ in this, since you already know the length of $u_\perp$ is the same as the equal lengths of $v_\perp,v_x.$ Finally the angle $t$ in the above polar form is taken to be called $\theta$ to match the diagram.
A: Per @Jean-Claude Arbaut's suggestion, 
$$ 
\left[\begin{array}{rrr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]
\left[\begin{array}{rrr}
v_\perp^1 \\
v_\perp^2
\end{array}\right] = 
\left[\begin{array}{rrr}
v_\perp^1 \cos \theta  - v_\perp^2 \sin \theta \\
v_\perp^2 \cos \theta  + v_\perp^1 \sin \theta  
\end{array}\right]
$$
$$ =
\cos \theta
\left[\begin{array}{rrr}
v_\perp^1  \\
v_\perp^2   
\end{array}\right] 
+ 
\sin \theta
\left[\begin{array}{rrr}
 - v_\perp^2  \\
 + v_\perp^1  
\end{array}\right] 
$$
Since $v_x$ is 90 degree rotated form of $v_\perp$, and that rotation looks like this, 
$$ 
\left[\begin{array}{rrr}
0 & -1 \\ 1 & 0
\end{array}\right]
\left[\begin{array}{rrr}
v_\perp^1 \\
v_\perp^2
\end{array}\right] = 
\left[\begin{array}{rrr}
-v_\perp^2 \\
v_\perp^1
\end{array}\right]
$$
RHS looks like the term after $\sin \theta$, so we can substitute $v_x$ there, then 
$$ = \cos \theta v_\perp + \sin \theta v_x $$
