How is $\sin^2(x)+\cos^2(x)=1$ where $x$ is an obtuse angle? Ok, so I know that $\sin^2(x)+\cos^2(x)=1$ for all angles. If $x$ is an acute angle in a right angled triangle it's a straightforward proof. 
But what about if $x$ is obtuse ? How do I mathematically prove it plus get a visual analysis of the same so you can use it in $2D$ or $3D$ geometry using the co-ordinate axes.
Edit: I know the proofs in which they show a unit circle. But I thought the obtuse angles were only designed to incorporate the sign. 
For example say a point on the unit circle is (-0.5,0.866). This shows up for x=120. However, we do the calculations for the acute angle from the negative x axis and then just put sign for sin(x) or cos(x) if x is any angle so that when we resubstitue x=r*sin(x) we get the polarity of x because sin(x) can be positive or negative but r is always considered postive.
r is the distance from the origin.
But what if we have a triangle in space which has coordinates say (1,1),(1,5) and (-2,8). How do we use trigonometry in those cases ? Since the angle made with the x axis for the above points doesn't matter since the angle between the linea joining those points are something totally different

A complex analysis reasoning:
Let
$$f(z)=\sin^2(z)+\cos^2(z)$$
$$g(z)=1$$
Since $f=g$ for all $z\in[0,\frac{\pi}2]$ and $f,g$ are holomorphic on $\mathbb C$, $f=g$ for all $z\in\mathbb C$ by identity theorem.
