Showing that $\left(1-2p^{2}\sin^2{\omega}\right)^2 + p^{2}\sin^{2} w\leq 1$ implies $4p^2 \leq 4$

Question

If $\xi = 1-2p^{2}\sin^{2} \omega -ip\sin w$, then the modulus of $\xi$ is

$$|\xi|^{2} = \left(1-2p^{2}\sin^2{\omega}\right)^2 + p^{2}\sin^{2} w$$

For $|\xi|^2 \leq 1$,

$$4p^2 \leq {4\sin^2{\omega \over 2}-\sin^2\omega \over \sin^4{\omega \over 2}}= 4$$

I am not sure how to get the second line. Can anyone show me? I have expanded out the bracket, etc, but do not get this expression.

Answer 1

Since $$1\geq 1-4p^2\sin^2{\omega \over 2}+4p^4\sin^4{\omega \over 2}+p^2\sin^2\omega$$

we have $$0\geq p^2(-4\sin^2{\omega \over 2}+4p^2\sin^4{\omega \over 2}+\sin^2\omega)$$

so $$0\geq -4\sin^2{\omega \over 2}+4p^2\sin^4{\omega \over 2}+\sin^2\omega$$

so $$ 4\sin^2{\omega \over 2}-\sin^2\omega\geq 4p^2\sin^4{\omega \over 2}$$

so $$ {4\sin^2{\omega \over 2}-\sin^2\omega \over \sin^4{\omega \over 2}}\geq 4p^2$$