## Precalculus (10th Edition)

Recall: (1) $\cos{(\alpha+\beta)}=\cos{(\alpha)}\cos{(\beta)}-\sin{(\alpha)}\sin{(\beta)}$ (2) $\cos{\frac{3\pi}{2}}=0$ and $\sin{\frac{3\pi}{2}}=-1$ Hence, $\cos{(\frac{3\pi}{2}+\theta)}\\=\cos{(\frac{3\pi}{2})}\cos{(\theta)}-\sin{(\frac{3\pi}{2})}\sin{(\theta)}\\=0\cos{(\theta)}-(-1)\cdot \sin{(\theta)}\\ =\sin{(\theta)}$