Different Law of Cosines using Sine instead: $c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$ Playing around with Trig and the Law of Cosines (LoC), I came up with this formula given a triangle with sides $a$, $b$, $c$ where we are given $a$, $b$ and angle $\theta$ between them:
$$c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$$
Far from me the idea that I could've stumbled onto something no one's ever derived before, but I've never seen this formula and was just curious whether it has a name or is never considered because it offers no advantage over the LoC (needing the same amount of initial information) and is slightly more complicated.
Also, is my proof correct?

Here's my work; here I use $C$ for the angle:
$$c^2 = x^2 + h^2$$
$$h = a \sin(C)$$ 
$$h^2 = a^2 \sin^2(C)$$
$$x = b - (b-x)$$
$$(b-x) = \sqrt{a^2 - h^2} = \sqrt{a^2 - a^2 \sin^2(C)} = \sqrt{a^2 (1-\sin^2(C))}$$
$$x = b-a\sqrt{1-\sin^2(C)}$$
$$x^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C))$$
Therefore:
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C)) + a^2 \sin^2(C)$$
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C) + \sin^2(C))$$
$$c^2 = a^2 + b^2 - 2ab \sqrt{1-\sin^2(C)}$$

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.
