<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] Really this just boils down to the identity $$\cos^2{x} = 1 - \sin^2{x}$$
So no, your formula is really no different from the Law of cosines. Your proof looks fine however. 
Having believed I found a new trigonometric law once (and being incorrect), my advice would be to always exhaust your result for any known identities to see if you simply have another representation of a known law or expression. Resources like Wolfram Alpha can help for easier cases. 
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