<s>[INST] $|\sin\phi_{n}(x)| = \sqrt{\frac{1}{2} - \dfrac{1}{2[1+ (\omega_{n}a(x))^{2}]^{1/2}}}$ Let a sequence of real numbers $\omega_{n}$ with $\omega_{n} \to \infty$ with $a(\cdot) \in L^{\infty}(0,1)$, $a(x) > 0$ for all $x \in (0,1]$ and $a(0) = 0$.
Let $\phi_{n}(x) = -\frac{1}{2}\text{arg}[1+i\omega_{n}a(x)]$, then
$$
|\sin\phi_{n}(x)| = \sqrt{\frac{1}{2} - \dfrac{1}{2[1+ (\omega_{n}a(x))^{2}]^{1/2}}}
$$
I am thinking that the author is using $\sin(\text{arg}(z)) = \dfrac{\text{Im}(z)}{|z|}$, but I do not understand.
 [/INST] We need the formulas $\arg(1+iy)=\arctan y$, $\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$ and $\cos\arctan x=\frac{1}{\sqrt{1+x^2}}$. Then,
\begin{align}
|\sin\phi_n (x)|&=|\sin(-\frac{1}{2}\arctan(w_na(x))|\\
&=\left|-\sin\left(\frac{\arctan(w_na(x))}{2}\right)\right|\\
&=\left|\sin\left(\frac{\arctan(w_na(x))}{2}\right)\right|\\
&=\sqrt{\frac{1-\cos\arctan(w_na(x))}{2}}\\
&=\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{1+(w_n a(x))^2}}}\\
\end{align}
</s>