Behauptung:	\
\cos \varphi = \sin \psi	\
\mbox{mit} \ \psi = (90^0 - \varphi)

P = P(x/y)

=(\cos \varphi / \sin \varphi)

ψ = 90° - φ

Die \ Dreiecke \ \Delta PDO \ und \ \Delta GOP'	  \
sind \ zueinander \ kongruent : wsw	\
\Rightarrow \cos \varphi = \sin \psi = \sin (90^0-\varphi)

Frage:	\
Gilt \ auch \ \sin \varphi = \cos (90^0 - \varphi) \ ?

O = Intersect(xAxis, yAxis)
Einheitskreis = Circle(O, 1)
B = (-1.8390974733399892, -0.7358162020793699)
f = Ray(O, B)
C = Intersect(Einheitskreis, xAxis, 2)
φ = Angle(C, O, B)
P = Intersect(Einheitskreis, f)
g = Line(P, xAxis)
h = Line(P, yAxis)
A = Intersect(h, xAxis)
cosφ = Vector(O, A)
D = Intersect(g, yAxis)
sinφ = Vector(O, D)
P' = Rotate(P, (-φ) + 90° - φ, O)
i = Ray(O, P')
E_1: (0, 1)
ψ = Angle(P', O, E_1)
j = Line(P', xAxis)
k = Line(P', yAxis)
E = Intersect(Einheitskreis, yAxis, 2)
G = Intersect(j, yAxis)
sinψ = Vector(O, G)
d1, d, p, o = Polygon(P, O, D)
d2, o_1, p', g_1 = Polygon(P', G, O)