The period of \(\dfrac{|\sin 4x|+|\cos 4x|}{|\sin 4x-\cos 4x|+|\sin 4x+\cos 4x|}\) is:
\(\dfrac{\pi}{8}\)
Let \(u=4x\). Using the identity \(|p-q|+|p+q|=2\max(|p|,|q|)\) with \(p=\sin u\), \(q=\cos u\), the denominator becomes \(|\sin u-\cos u|+|\sin u+\cos u|=2\max(|\sin u|,|\cos u|)\).
So the expression becomes \(g(u)=\dfrac{|\sin u|+|\cos u|}{2\max(|\sin u|,|\cos u|)}\).
Check \(g\left(u+\dfrac{\pi}{2}\right)\): since \(\sin\left(u+\dfrac{\pi}{2}\right)=\cos u\) and \(\cos\left(u+\dfrac{\pi}{2}\right)=-\sin u\), both the numerator (sum of absolute values) and the denominator (max of absolute values) are unchanged when \(\sin u\) and \(\cos u\) swap places, so \(g\left(u+\dfrac{\pi}{2}\right)=g(u)\) for all u. Hence \(\dfrac{\pi}{2}\) is a period in u.
Check whether a smaller period \(\dfrac{\pi}{4}\) works: at \(u=0\), \(g(0)=\dfrac{0+1}{2\cdot1}=\dfrac12\); at \(u=\dfrac{\pi}{4}\), \(g\left(\dfrac{\pi}{4}\right)=\dfrac{\frac{\sqrt2}{2}+\frac{\sqrt2}{2}}{2\cdot\frac{\sqrt2}{2}}=1\). Since \(g(0)\ne g(\pi/4)\), \(\pi/4\) is not a period, so \(\pi/2\) is the fundamental period in u.
Since \(u=4x\), the period in x is \(\dfrac{\pi/2}{4}=\dfrac{\pi}{8}\).
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