Context
It's pretty easy to prove said identity for angles smaller than 90 degrees, because we can use a right-angled triangle, and the result falls out of the definitions of $\sin$ and $\cos$ inside the triangle.
What I'd like to do, is prove it more generally, but I'm unsure of how I can visualize $\frac\pi2 - \theta$ if $\theta > \frac\pi2$.
As an example, if we draw an angle in the 4th quadrant, we get the angles $\frac\pi2, \quad 2\pi - \theta, \quad x - \frac{3\pi}2$.
Question
What's a nice visual proof of $\sin(\frac\pi2 - \theta) = \cos(\theta)$ for angles larger than $\frac\pi2$?
Caveat
I'm aware of proofs that involve algebraically deducing it using other identities like angle sums etc. I'm specifically seeking direct, visual proofs for this one.
Hint:
In polar coordinates, $\theta\mapsto\frac\pi 2-\theta$ is the composition of a symmetry w.r.t. the abscissæ axis by a rotation of $\frac\pi 2$.
Other interpretation: Points on the unit circle with polar angles $\theta$ and $\frac\pi 2-\theta$ are symmetric w.r.t. the first bissectrix.