Talk:Polarization correction

Angle		Adjacent	Opposite	Hypotenuse	Sine = O/H	Cosine = A/H	Tangent = O/A
$\chi$	Azimuth on detector (relative to $q_{z}$ axis)	$z$	$x$	$r={\sqrt {x^{2}+z^{2}}}$	$\sin \chi ={\frac {x}{r}}$	$\cos \chi ={\frac {z}{r}}$	$\tan \chi ={\frac {x}{z}}$
$2\theta$	Full scattering angle (between incident beam and scattering)	$y$	$r={\sqrt {x^{2}+z^{2}}}$	$R={\sqrt {x^{2}+y^{2}+z^{2}}}$	$\sin 2\theta ={\frac {r}{R}}$	$\cos 2\theta ={\frac {y}{R}}$	$\tan 2\theta ={\frac {r}{y}}$
$\gamma$	In-plane angle	$y$	$x$	$h={\sqrt {x^{2}+y^{2}}}$	$\sin \gamma ={\frac {x}{h}}$	$\cos \gamma ={\frac {y}{h}}$	$\tan \gamma ={\frac {x}{y}}$
$\delta$	Elevation angle	$h={\sqrt {x^{2}+y^{2}}}$	$z$	$R={\sqrt {x^{2}+y^{2}+z^{2}}}$	$\sin \delta ={\frac {z}{R}}$	$\cos \delta ={\frac {h}{R}}$	$\tan \delta ={\frac {z}{h}}$

So:

${\begin{alignedat}{2}\tan \gamma &={\frac {x}{y}}\\&={\frac {r\sin \chi }{r/\tan 2\theta }}\\&=\tan 2\theta \sin \chi \\\gamma &=\tan ^{-1}\left[\tan 2\theta \sin \chi \right]\end{alignedat}}$

and: TBD

and so:

${\begin{alignedat}{2}P_{h}&=1-\cos ^{2}\delta \sin ^{2}\gamma \\&=1-\cos ^{2}\delta {\frac {\left[\tan 2\theta \sin \chi \right]^{2}}{\left[\tan 2\theta \sin \chi \right]^{2}+1}}\\\end{alignedat}}$