Pseudo-pythagorean theorem Pythagoras' theorem is a special case of the Cosine theorem for a angle of $90°$. But also for an angle of 60° and 120°, "aesthetical" special cases derive:
$c^2=a^2+b^2\pm ab$
First question:
Are there further angle $x°$ with a rational number $x$, so that $\cos x$ is rational as well, thus creating "aesthetical" special cases?
Second question:
Does anybody know some internet sources to the equivalent of pythagorean triplets (Integers $a,b,c$, so that $c^2=a^2+b^2\pm ab$)?

To find all integer triples $a,b,c$ for which $c^2=a^2+b^2\pm ab$, it's enough to consider only $c^2=a^2+b^2+ab$ because changing the sign of $a$ or $b$ gives a solution to the other equation and vice versa.
We'll only look for primitive solutions, i.e, satisfying $\gcd(a,b,c)=1$. Then at least one of $a$ or $b$ is odd, say it's $b$.
Note that the equation can be rewritten
$$4c^2=(2a+b)^2+3b^2.$$
Set $d=2a+b$, so $d$ is odd. We have
$$3b^2=(2c-d)(2c+d)$$
where $\gcd(2c-d,2c+d)=1$ because of the primitivity. There are two cases: $2c-d=3u^2$ and $2c+d=v^2$, or $2c-d=v^2$ and $2c+d=3u^2$. In any case we have $\gcd(u,v)=1$, $3\nmid v$ and $b=uv$, with $u$ and $v$ odd.
The first case gives $c=\frac{v^2+3u^2}4$ and $d=\frac{v^2-3u^2}2$ from which $a=\frac{v^2-2uv-3u^2}4$.
The second case yields $c=\frac{v^2+3u^2}4$ and $d=\frac{3u^2-v^2}2$ from which $a=\frac{3u^2-2uv-v^2}4$.
In both cases $b=uv$.
As said before, we chose $b$ to be odd. This means we can change the values found for $a$ and $b$. (We could be introducing doubles here in the case where $a$ and $b$ are both odd. In fact, I'm sure we do.)
To obtain all values (including the non-primitive) we simply have to multiply by an integer constant $k=\gcd(a,b,c)$.
Example
Take coprime odd integers $u$ and $v$ with $3\nmid v$. Say $u=3$ and $v=5$. Then with the first case we obtain $a=-8$, $b=15$ and $c=13$. Indeed,
$$13^2=(-8)^2+15^2+(-8)\cdot15.$$
The second case gives the same values for $b$ and $c$, but $a=-7$. Again,
$$13^2=(-7)^2+15^2+(-7)\cdot15.$$
