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 $c^2=a^2+b^2\pm ab$)?
 A: If $\cos\frac {2m\pi}n$ is rational with $\gcd(m,n)=1$, then the primitive $n$th root of unity $\zeta=\cos \frac{2m\pi}{n}+i\sin\frac{2m\pi}n$ is a root of the rational polynomial $X^2-2\cos \frac{2m\pi}{n}X+1$. On the other hand, we know that $\zeta$ is a root of $X^n-1$ and any factorization of this over the rationals can be resaled to a factorization over the integers. Since $X^2-2\cos \frac{2m\pi}{n}X+1$ is monic this implies that $2\cos\frac{2m\pi }{n}$ is already an integer of absolute value $\le 2$.
This leads to the cases


*

*$c^2=a^2+b^2+2ab$ for $\gamma = \pi$

*$c^2=a^2+b^2+ab$ for $\gamma = \frac 23\pi$ or $\gamma = \frac 43\pi$

*$c^2=a^2+b^2$ for $\gamma =\frac 12\pi$ or $\gamma = \frac 32\pi$

*$c^2=a^2+b^2-ab$ for $\gamma=\frac13\pi$ or $\gamma =\frac 53\pi$

*$c^2=a^2+b^2-2ab$ for $\gamma =0$


and that's all with $\gamma \in[0,2\pi)\cap \pi\mathbb Q$.

Regarding your second question: Pythagorean triples are best viewed as numbers $z=a+bi\in\mathbb Z[i]$ with norm $z\bar z$ a perfect square, which leads to a partitioning of the set of primes into those with $p\equiv -1\pmod 4$ (which can only occur as factor of $c$ if they are also factors of $a$ and $b$) and those with $p\equiv 1\pmod 4$ (which can be written as sums of squares and lead to primitive pythagorean triangles; for example $5=2^2+1^2=(2+i)(2-i)$ give us $\color{red}5^2=(2+i)^2(2-i)^2=(\color{red}3+\color{red}4i)(3-4i)$ and hence the most famous Pythagorean triangle) and the special prime $2$. Similarly, the numbers aou are after should be viewed as elements of $\mathbb Z[\omega]$ where $\omega=-\frac12+\frac i2\sqrt 3$. It turns out that one gets a similar partitioning of the primes, this time based on the remainders modulo $3$. Those are very interesting number theoretic questions indeed.
A: Regarding your first question:
The law of cosines states that
$$c^2=a^2+b^2-2ab\cos(\gamma)$$
If $r$ is any rational between $-1$ and $1$ you can set $\gamma=\arccos(r)$, therefore giving you
$$c^2=a^2+b^2-2abr$$
However the angle $\gamma$ is not a rational multiple of $\pi$ unless $r=0,\pm  1/2,\pm 1$. The latter is known as Niven's theorem.
A: 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,
$$13^2=(-7)^2+15^2+(-7)\cdot15.$$
A: This is a supplement of the answer by @barto.
The problem with the analysis there is that, it is not obvious that $\gcd(2c-d,2c+d)=1.$ So let $m=\gcd(2c-d,2c+d).$ Then we distinguish between two cases:  
I. $3\not\mid m$
By the equation $3b^2=(2c-d)(2c+d)$ we observe that $m\mid b$ (actually $m^2\mid b^2,$ hence $(b/m)^2\in\mathbb Z,$ hence $m\mid b$), from which we deduce that $m\mid\gcd(a,b,c)=1,$ thus $m=1.$  
II. $3\mid m$
Write $m=3n$ so that $$b^2=3n^2(\frac{2c-d}{m})(\frac{2c+d}{m}).$$ Hence $n^2\mid b^2.$ By the same token as in I. we conclude that $n\mid b.$ Write then $b=b'n$ to find:  $$(b')^2=3(\frac{2c-d}{m})(\frac{2c+d}{m}),$$ so $3\mid b'.$ This implies that $m=3n\mid b$ again. So $m\mid1,$ a contradiction.  
Therefore, with our assumptions, indeed it cannot happen that $\gcd(2c-d,2c+d)\not=1,$ but this is not so obvious, and is not a mere consequence of the primality of the solution.
Hope this clarifies some point.
