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What is the period of the function $f(x)= \sin^2 (x)$? Please explain in detail. I don't know anything about the periodicity of trigonometric function. How will I evaluate it?

Michael Rozenberg
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Sameer Sharma
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    The period of $\sin x$ is $2\pi$, here you would take only the positive part, so the period will be $\pi$ – Shailesh Jun 21 '17 at 11:01
  • "I don't know anything about the periodicity of trigonometric function..." Then start reading your mathematical scripts concerning it. If you're done and still don't understand things then this site can help. – drhab Jun 21 '17 at 11:05
  • see here: https://math.stackexchange.com/questions/164221/period-of-the-sum-product-of-two-functions – farruhota Jun 21 '17 at 11:13
  • Plot the two functions $g(x) = \sin(x)$ and $f(x) = \sin^2(x)$ on the same graph and look at them. That won't prove anything, but you will understand. – NickD Jun 21 '17 at 12:30
  • Squaring/cubing of pure sine wave leaves periodicity unchanged. – Narasimham Jun 21 '17 at 13:59

2 Answers2

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$$\sin^2x=\frac{1}{2}(1-\cos2x),$$ which gives the answer: $\pi$.

We can find a period by the definition.

Function $f$ is called periodic function if there are a number $T>0$, for which for all $x$ from the domain of $f$ we have $$f(x+T)=f(x).$$

If there exist the minimal $T$, about which we said in the definition, then $T$ is called the period of $f$.

We need to find a minimal $T>0$, for which $\sin^2(x+T)=\sin^2x$ for all real $x$ or $$\frac{1}{2}(1-\cos(2x+2T))=\frac{1}{2}(1-\cos2x)$$ or $$\cos(2x+2T)=\cos2x,$$which gives for all $x\in\mathbb R$:

$2x+2T=-2x+2\pi k$, where $k\in\mathbb Z$, which is impossible for all $x\in\mathbb R$ or

$2x+2T=2x+2\pi k$, where $k\in\mathbb Z$, which gives $T=\pi k$ for some $k$ and since $$\min_{k\in\mathbb Z}\{\pi k|\pi k>0\}=\pi,$$ we obtain $T=\pi$.

Michael Rozenberg
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$$f (x)=\sin^2 (x)=\frac {\tan^2 (x)}{1+\tan^2 (x)} $$

the period of $x\mapsto \tan (x) $ is known to be $\pi $. thus $\pi $ is the smallest positive $T $ real such that : $f (x+T)=f (x) $.

hamam_Abdallah
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