$$f(x)=\frac{1}{2}\left(\frac{|\sin x|}{\cos x}+\frac{|\cos x|}{\sin x}\right)$$

Then find it's period.

My teacher did this question with graphs. He considered the following intervals $0\to\frac{\pi}{2}$, $\frac{\pi}{2}\to\pi$, $\pi\to\frac{3\pi}{2}$ and $\frac{3\pi}{2}\to2\pi$

He then drew the graph in all the four quadrants and concluded that the period is $2\pi$. But I did it algebraically and find that it's period is something else. Let me show.

My attempt

If $f(x)$ is periodic, then $f(x+T)=f(x)$ where $T$ is the period $$\implies\frac{1}{2}\left(\frac{|\sin (x+T)|}{\cos (x+T)}+\frac{|\cos (x+T)|}{\sin (x+T)}\right)=\frac{1}{2}\left(\frac{|\sin x|}{\cos x}+\frac{|\cos x|}{\sin x}\right)$$ Putting $x=0$ on both sides, we get $$\frac{1}{2}\left(\frac{|\sin T|}{\cos T}+\frac{|\cos T|}{\sin T}\right)=\frac{1}{2}\left(\frac{|\sin 0|}{\cos 0}+\frac{|\cos 0|}{\sin 0}\right)$$ $$\implies \frac{|\sin T|}{\cos T}+\frac{|\cos T|}{\sin T}=\frac{\left|\cos\frac{\pi}{2}\right|}{\sin\frac{\pi}{2}}+\frac{\left|\sin\frac{\pi}{2}\right|}{\cos\frac{\pi}{2}}$$ $$\implies T=\frac{\pi}{2}$$ Hence the period is $\frac{\pi}{2}$

Now who is right and who is wrong and why$?$

Any help is greatly appreciated.

  • $\begingroup$ $$\frac{\cos 0}{\sin 0}=\frac{1}0$$ $\endgroup$
    – MathFail
    Commented May 29, 2023 at 11:21

1 Answer 1


The way followed from your teacher is fine, since both $\sin x$ and $\cos x$ are periodic with period $2\pi$ the given function has $2\pi$ as a period, and showing that there is not a smaller period, by consideration on the graph, complete the proof.

What you have tried to show is that $f(0+T)=f(0)$ but this condition doesn't suffice in general to determine the period. Moreover, as already noticed, the expression you have found is not well defined, therefore this is neither a proof for that fact.


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