Why is this inequality true? Trigonometry \begin{align}\lvert \cos\theta(\cos\theta -i\sin\theta)\rvert&amp;\leq \lvert \cos\theta -i\sin\theta\rvert\\&amp;=\sqrt{\cos^2\theta+\sin^2\theta}=1
\end{align}
I know $0\leq\lvert \cos\theta \rvert\leq 1$, but why can we just remove $\lvert \cos \theta \rvert$ on the right hand side of the inequality?
Note that since $\vert\cos(x)\vert\le1$ and $\vert x_1x_2\vert=\vert x_1\vert\cdot\vert x_2\vert$ (you should verify this yourself), then $$\left\vert\cos(x)(\cos(x)-i\sin(x))\right\vert=\left\vert\cos(x)\right\vert\cdot\left\vert(\cos(x)-i\sin(x))\right\vert\le1\cdot\left\vert(\cos(x)-i\sin(x))\right\vert=\sqrt{\cos^2(x)+\sin^2(x)}=1$$
It is generally true that $|ab|=|a||b|$, so that in this case: 
$|cos(\theta)(cos(\theta)-isin(\theta))|=|cos(\theta)|(cos(\theta)-isin(\theta))|\leq|(cos(\theta)-isin(\theta))| $.