What is the fundamental period of $f(x)=\frac{|\sin x|+|\cos x|}{|\sin x -\cos x|}$

Question

I tried it.
$f(x)=\frac{|\sin x|+|\cos x|}{|\sin x -\cos x|}=\frac{|\sin x|+|\cos x|}{\sqrt 2|\sin (x-\frac{\pi}{4})|}$.

I know that the period of $|\sin x|+|\cos x|$ is $\frac{\pi}{2}$.But i cannot find the period of whole function.Please help me.

@Math-fun: that does not prove $\pi$ is the fundamental period. Why not $\frac{\pi}{5}$ or $\frac{\pi}{3}$? — Jack D'Aurizio, Sep 25 '15 at 12:31
@JackD'Aurizio I did not mean to give a proof :-) I simply gave the OP the answer. I thought this might help and direct his thoughts towards specific candidates for the fund. period. Many thanks for the comment, though :-) — Math-fun, Sep 25 '15 at 13:55
@Math-fun: all right, I am fine with that. Have a nice day ;) — Jack D'Aurizio, Sep 25 '15 at 15:57

score 2 · Answer 1 · answered Sep 25 '15 at 12:01

Just look at the solutions of $f(x)=f(0)=1$.

$f(x)=1$ implies $\left|\sin x\right|+\left|\cos x\right|=\left|\sin x-\cos x\right|$, but the triangle inequality gives $\left|\sin x-\cos x\right|<\left|\sin x\right|+\left|\cos x\right|$ over $\left(0,\frac{\pi}{2}\right)$. That gives that the fundamental period is $\geq \frac{\pi}{2}$. By direct inspection, $\frac{\pi}{2}$ is not a period, but $\pi$ is, hence $\pi$ is the fundamental period.

Also notice that $f(x)$ is constant over $\left[\frac{\pi}{2},\pi\right]$.

What is the fundamental period of $f(x)=\frac{|\sin x|+|\cos x|}{|\sin x -\cos x|}$

1 Answers1