# Showing that ( 1 &#x2212;<!-- - --> 2 p

Showing that ${\left(1-2{p}^{2}{\mathrm{sin}}^{2}\omega \right)}^{2}+{p}^{2}{\mathrm{sin}}^{2}w\le 1$ implies $4{p}^{2}\le 4$
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Since
$1\ge 1-4{p}^{2}{\mathrm{sin}}^{2}\frac{\omega }{2}+4{p}^{4}{\mathrm{sin}}^{4}\frac{\omega }{2}+{p}^{2}{\mathrm{sin}}^{2}\omega$
we have
$0\ge {p}^{2}\left(-4{\mathrm{sin}}^{2}\frac{\omega }{2}+4{p}^{2}{\mathrm{sin}}^{4}\frac{\omega }{2}+{\mathrm{sin}}^{2}\omega \right)$
so
$0\ge -4{\mathrm{sin}}^{2}\frac{\omega }{2}+4{p}^{2}{\mathrm{sin}}^{4}\frac{\omega }{2}+{\mathrm{sin}}^{2}\omega$
so
$4{\mathrm{sin}}^{2}\frac{\omega }{2}-{\mathrm{sin}}^{2}\omega \ge 4{p}^{2}{\mathrm{sin}}^{4}\frac{\omega }{2}$
so
$\frac{4{\mathrm{sin}}^{2}\frac{\omega }{2}-{\mathrm{sin}}^{2}\omega }{{\mathrm{sin}}^{4}\frac{\omega }{2}}\ge 4{p}^{2}$