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I need to find the period of $\frac{\cos 3x}{1+\cos2x}$. I know that $\cos3x$ has the period $\frac{2\pi}{3}$ and $\cos2x$ has the period $\pi$, but how do I find the period of that function?

Davide Giraudo
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Ghost
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1 Answers1

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The expression is equivalent to $$2\cos x-\frac 32\sec x$$ and both of these functions have a period of $2\pi$ so the period is $2\pi$

David Quinn
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  • Strictly speaking, only a period is $2\pi$. In principle, the sum of two functions with period $2\pi$ may happen to have period $\frac{2\pi}n$ for some natural $n$. – Hagen von Eitzen Apr 09 '17 at 10:55
  • @HagenvonEitzen I understand the OP is looking for the smallest period, usually just referred to as the period, which in this case is indeed $2\pi$ – David Quinn Apr 09 '17 at 11:17
  • @DavidQuinn An example of what I think he means to say is, if $f(x) = \sin{x}+\sin{2x}$ has period $2\pi$, $g(x) = - \sin{x}$ has period $2 \pi$, but $f+g$ has period $\pi$. So, what you have proved is that $2\pi$ is "a" period, not the smallest period. – IamThat Apr 09 '17 at 12:38
  • @bat_of_doom but $\cos x$ and $\sec x$ both have the same period so your example does not apply in this case. – David Quinn Apr 09 '17 at 13:23
  • @DavidQuinn Indeed, it doesn't apply. That needs to be proved. The statement is false even for periodic functions having the same period, for example, $\cos x + (-\cos x +1)$ is not even periodic. Anyway, check out the proposed edit. I have tried to add the proof of that statement too, using the definition of periodicity. – IamThat Apr 10 '17 at 06:01