$26 \cdot 3$ If a moving point $P(x, y)$ rotates counterclockwise on the unit circle with a constant angular velocity $\omega$, then the motion of point $Q\left(-2 x y, y^{2}-x^{2}\right)$ is
(A) rotating clockwise on the unit circle with angular velocity $\omega$.
(B) rotating counterclockwise on the unit circle with angular velocity $\omega$.
(C) rotating clockwise on the unit circle with angular velocity $2 \omega$.
(D) rotating counterclockwise on the unit circle with angular velocity $2 \omega$.
(China High School Mathematics League, 1984)
[Solution] $\because$ A moving point $P(x, y)$ moves counterclockwise on the unit circle with a constant angular velocity $\omega$,
$\therefore$ we can set $x=\cos \omega t, y=\sin \omega t$,
then $-2 x y=-2 \cos \omega t \sin \omega t=-\sin 2 \omega t=\cos \left(270^{\circ}-2 \omega t\right)$,
$y^{2}-x^{2}=\sin ^{2} \omega t-\cos ^{2} \omega t=-\cos 2 \omega t=\sin \left(270^{\circ}-2 \omega t\right)$.
This indicates that point $Q$ starts from $(0,-1)$, moves clockwise on the unit circle, and has an angular velocity of $2 \omega$.
Therefore, the correct choice is $(C)$.