<s>[INST] 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:
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C)) + a^2 \sin^2(C)$$
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C) + \sin^2(C))$$
$$c^2 = a^2 + b^2 - 2ab \sqrt{1-\sin^2(C)}$$
 [/INST] 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.
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