Answer
$$\frac{d}{d\omega}(a\cos^2\pi\omega+b\sin^2\pi\omega)=(b\pi-a\pi)\sin2\pi\omega$$
Work Step by Step
$$\frac{d}{d\omega}(a\cos^2\pi\omega+b\sin^2\pi\omega)=a\cdot2\cos\pi\omega\frac{d}{d\omega}(\cos\pi\omega)+b\cdot2\sin\pi\omega\frac{d}{d\omega}(\sin\pi\omega)=
2a\cos\pi\omega\cdot(-\sin\pi\omega)\frac{d}{d\omega}(\pi\omega)+2b\sin\pi\omega\cdot\cos\pi\omega\frac{d}{d\omega}(\pi\omega)=
-2a\cos\pi\omega\sin\pi\omega\cdot\pi+2b\cos\pi\omega\sin\pi\omega\cdot\pi=
(b\pi-a\pi)\cdot 2\cos\pi\omega\sin\pi\omega=(b\pi-a\pi)\sin2\pi\omega$$
