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:
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C)) + a^2 \sin^2(C)$$
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C) + \sin^2(C))$$
$$c^2 = a^2 + b^2 - 2ab \sqrt{1-\sin^2(C)}$$
 A: 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.
A: 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.
A: 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. 
