Q:

Flawed law of cosines application

I've been studying basic trigonometry since I forgot most of it, and my problem is with the law of cosines.
I already understood how I can get the cosines law in the simplified way to get sides from an angle:
$$a² = b²+c²-2bc\cos(a)$$
To test it I imagined an isosceles triangle with two sides 1, and the angle between those sides as 30 degrees. Here is my triangle
And with that I could conclude that the missing side of that triangle was sqrt(2-2 cos(30)), yet unsure if it was right, so I checked with the law of Sines, and then with wolfram to be pretty sure it was right.
Meanwhile I tried to reverse it, or in other words: Get the 30 degree angle from the sides I had.
Since I was confused with some other explanations online which gave different formulas for the same thing, I tried to deduce mine:
$$a² = b²+c²-2bc\cos(A) <=> b²+c²-a² = 2bc\cos(A) <=>
 \frac{b²+c²-a²}  {2bc} = \cos(A)$$
I looked at it for a minute or two, and it seems fine, although I spent too much time without doing math and I wouldn't be surprised if it is wrong somehow.
The problem is that I can't get the cosine of the angle (30º) with it.
$$\frac{1²+1²-\sqrt{2-2\cos30}^2}{2}$$
 Simplifies to: $$\frac{-2\cos30}{2}$$
Which is the cosine of 150º, the symmetrical angle(relating to the axis of cosines) of 30º, which leads me to think that the mistake is with the signals. It returns -0.86 instead of 0.86. Yet i can't find the problem anywhere, and after an hour or so I found this formula for isosceles triangles which works just fine.
What's my mistake? Thanks

A:

$$\frac{1²+1²-\sqrt{2-2\cos30}^2}{2}$$
$$=\frac{2-(2-2\cos 30^\circ)}{2}$$
$$=\frac{2-2+2\cos 30^\circ}{2}$$
Do you see your mistake?

