Answer
$$\frac{\partial}{\partial r}\sin(r\cos\theta)=\cos\theta\cos(r\cos\theta).$$
$$\frac{\partial}{\partial \theta}\sin(r\cos\theta)=-r\sin\theta\cos(r\cos\theta).$$
Work Step by Step
Regard $\theta$ as a constant and differentiate with respect to $r$:
$$\frac{\partial}{\partial r}\sin(r\cos\theta)=\cos(r\cos\theta)\frac{\partial}{\partial r}(r\cos\theta)=\cos(r\cos\theta)\cos\theta\frac{\partial}{\partial r}r=\cos\theta\cos(r\cos\theta).$$
Regard $r$ as a constant and differentiate with respect to $\theta$:
$$\frac{\partial}{\partial \theta}\sin(r\cos\theta)=\cos(r\cos\theta)\frac{\partial}{\partial \theta}(r\cos\theta)=\cos(r\cos\theta)r\frac{\partial}{\partial\theta}(\cos\theta)=-r\sin\theta\cos(r\cos\theta).$$
