Different Law of Cosines using Sine instead: $c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$

Yours is same as the cosine rule. Recall that $$c^2 = a^2+b^2-2ab\cos(C)$$ Now note that if $C$ is acute, we then have that $\cos(C) = \sqrt{1-\sin^2(C)}$. Hence, we obtain $$c^2 = a^2+b^2-2ab\sqrt{1-\sin^2(C)}$$ Your proof is fine, though note that if $\angle{C}$ were to be obtuse, then writing $x$ as $b+(x-b)$ would be the right way to go about.

Due to the nature of the square root I think this equation you have is less useful than the 'usual' Law of Cosines. We have that $$c^2 = a^2+b^2-2ab\cos(\theta_c)$$ where $\theta_c$ is the angle opposite of the triangle side $c$. By the Pythagorean Theorem we also know $$\sin^2(\theta_c)+\cos^2(\theta_c) = 1$$ and solving for $\cos(\theta_c)$ gets us $\cos(\theta_c) = \sqrt{1-\sin^2(\theta_c)}$, hence in a single step we can get to the equation you have, $$c^2 = a^2+b^2-2ab\sqrt{1-\sin^2(\theta_c)}$$ However we also know that $\cos(\theta_c)$ will be negative when $\pi/2 < \theta_c <3\pi/2$, while $\sqrt{1-\sin^2(\theta_c)}$ will always be non-negative. So you would need to make cases for your equation to be accurate. $$c^2 = a^2+b^2-2ab\sqrt{1-\sin^2(\theta_c)} \quad \text{when} \space -\pi/2 \leq \theta_c \leq \pi/2$$ and $$c^2 = a^2+b^2+2ab\sqrt{1-\sin^2(\theta_c)} \quad \text{when} \space \pi/2 < \theta_c <3\pi/2$$ At this point it seems more reasonable to avoid cases and use the 'usual' Law of Cosines.