$\sin(x+iy)=\sin x\cos(iy)+\cos x \sin(iy)=\sin x\cosh y + i \cos x\ sinh y$
$|\sin(z)|=|\sin(x+iy)|=\sqrt{\sin^2x\cosh^2y+\cos^2x\sinh^2y}=\sqrt{\cosh^2y-cos^2x}$ as $sin^2A+cos^2A=1$ and $cosh^2B-sinh^2B=1$
Now if we consider points on the two lines $x=±(2m+1)\frac{\pi}{2}$,
$\cos x=0$ so, $|\sin(z)|=|\sin(x+iy)|=|\cosh y|=|\frac{e^y+e^{-y}}{2}|$ at $x=±(2m+1)\frac{\pi}{2}$.
Now for the positive real value of $d$, $d+\frac{1}{d}=(\sqrt d-\frac{1}{\sqrt d})^2+2≥2$
$|\sin(z)|=|\sin(x+iy)|≥1$ for any real $y$ at $x=±(2m+1)\frac{\pi}{2}$
Now if we consider points on the two lines $y=±(2m+1)\frac{\pi}{2}$,
$|\sin(z)|=|\sin(x+iy)|=\sqrt{\cosh^2y-cos^2x}=\sqrt{\sinh^2y+sin^2x}$
Now, $e^p-e^{-p} > e^q-e^{-q}$ iff $(e^p-e^q)(e^{p+q}-1)>0$ which is true if $p>q>0$
So, $(e^r-e^{-r})^2 > (e-e^{-1})^2 $ if $r^2>1$
$sinh^2y=(\frac{e^y-e^{-y}}{2})^2>(\frac{e-e^{-1}}{2})^2$ if $y=±(2m+1)\frac{\pi}{2}$,
But $e-e^{-1}>2$ as $e^2-2e-1>0$ as $(e-1)>\sqrt2$
So, $sinh^2y>1$ if $y=±(2m+1)\frac{\pi}{2}$,
and $|\sin(z)|=|\sin(x+iy)|=\sqrt{\cosh^2y-cos^2x}$ $=\sqrt{\sinh^2y+sin^2x}>\sqrt{1+sin^2x}>1$ if $y=±(2m+1)\frac{\pi}{2}$,