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

P = P(x/y)

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

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

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

O = Intersect(xAxis, yAxis)
Einheitskreis = Circle(O, 1)
B = (-1.532482497417815, 0.8600920418194049)
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, -90°, O)
β = Angle(P', O, P)
i = Ray(O, P')
j = Line(P', xAxis)
k = Line(P', yAxis)
ψ = Angle(C, O, P')
E = Intersect(Einheitskreis, yAxis, 2)
F = Point(xAxis)
cosψ = Vector(O, F)
d1, d, p, o = Polygon(P, O, D)
d2, p', o_1, f_1 = Polygon(O, F, P')