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In my textbook, it is stated that the period of $\sin(2x)$ is $\pi$. The author justifies this using a mathematical statement which I cannot understand. He writes that, since $\sin(2x) = \sin(2x+2\pi) = \sin(2(x+\pi))$ the period of $\sin(2x)$ is $\pi$.

Though my intuition tells me that the period of $\sin(2x)$ is $\pi$, I just cannot understand this reasoning. To me the period of $\sin(2x)$ appears to be to $2\pi$ since $\sin(2x)=\sin(2x+2\pi)$. I would be very thankful if someone could explain this reasoning to me.

Sahan Manodya
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MrAP
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    You are confusing the meaning of the function $\sin(2x)$. You can let $f(x) = \sin(2x)$ to resolve your confusion. Then, you would see that the last statement you made is actually $f(x)=f(x+\pi)$. – Kenny Lau May 03 '17 at 18:05
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    The period of a function $f:\mathbb{R}\to \mathbb{R}$ is the smallest positive $t$ such that $f(x+t) = f(x)$. In your example, $f(x) = \sin(2x)$ and $f(x + \pi) = f(x)$. – Darth Geek May 03 '17 at 18:06
  • $T$ is a period of $f$ if $f(x)=f(x+T)$ for all $x$. Let $f(x)=\sin 2x$ and $T=\pi$. $f(x+T)=\sin2(x+\pi)=\sin(2x+2\pi)=\sin2x$. So $\pi$ is a period. $2\pi$ is also a period as $f(x+2\pi)=\sin(2x+4\pi)=\sin 2x$. However, we usaually take the smallest possible period. – CY Aries May 03 '17 at 18:07
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    This boils down to: If an object spins twice as fast, it takes half as long to make one turn. – Semiclassical May 03 '17 at 18:08
  • @Semiclassical, I can understand this intuitively but it is the reasoning which is puzzling me. – MrAP May 03 '17 at 18:10
  • $\sin(x)$ is $\sin([x])$ and $\sin([x]+2\pi])=\sin([x+2\pi])=\sin[x]$ so the period is $2\pi$. because we stuck the 2\pi inside the []. $\sin(2x)$ is $\sin(2[x])$ and $\sin(2[x]) = \sin(2[x]+2\pi) = \sin(2[x + \pi])$ so the period is $\pi$ because it is $\pi$ (not $2\pi$) that ends up being stuck inside the []. – fleablood May 03 '17 at 18:47

5 Answers5

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The period of $\sin$ is $2\pi$; so $\sin (2x + 2\pi) = \sin (2x)$ for all $x$. On the other hand, we have $\sin (2x + 2\pi) = \sin (2(x+\pi))$ for all $x$. So $$ \sin (2x) = \sin (2(x+\pi)); $$ by definition the function $x \mapsto \sin (2x)$ has period $=\pi$.

Note that $x \mapsto \sin (2x)$ is a composite function; so it is not that obvious how to link the definition of periodic functions with the present case. I guess it could be this that confused you.

Yes
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  • How "by definition we proved that $x↦sin(2x)$ has period = $π$"? Are you referring to the definition of a periodic function? – MrAP May 03 '17 at 18:38
  • Indeed. I assumed you were given the most commonly seen definition. To avoid unnecessary misunderstandings I amended the sentence. – Yes May 03 '17 at 18:54
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If $f(x)$ has period $p$, then $f(ax)$ has period $p/a$.

Proof

Let $g(x)=f(ax)$.

$g(x)=f(ax)=f(ax+p)=f(ax+a\cdot \frac{p}{a})=f(a(x+p/a))=g(x+p/a)$

Ganesh Kumar
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By the definition of a period of a real function $f$: $$T=\inf_{t \in \mathbb R}\{t\mid \forall x\in \mathbb R:f(x)=f(x+t)\}$$ We can deduce from that the period of $f(x)=\sin(2x)$: $$T_{\sin(2x)}=\inf_{t \in \mathbb R}\{t\mid \forall x\in \mathbb R:\sin(2x)=\sin(2x+2t)\}=\pi$$

Arthur
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Alon Yariv
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  • I believe the thing you have defined is the basic or fundamental period. A period is any multiple of the basic period. – user Dec 04 '23 at 09:10
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I see this thread is old but to explain in simple terms for anyone still wondering (I myself didn't understand the above explanations but im sure they are great)

Sinx has a period of 2pi

Think of sine 2x as just 2x= 2pi

dividing both sides by 2 makes x=pi hence the period being pi

This doesn't prove anything like the above explanations but its a much easier (imo) way to remember how to find the period of a trig function graph

Egg
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The period of a function $f$ is (informally) the smallest value of $k$ (if any) so that $f(x + k) = f(x)$ for all $k$.

The period of $\sin()$ is $2\pi$ as you no doubt accept. We'll take that as a given.

$f(x) = \sin(2x)$ is a different function.

$f(x + \pi) = \sin (2(x+\pi)) = \sin (2x + 2\pi) = \sin (2x) = f(x)$. So the period of $f$ is $\pi$ or smaller. (It isn't smaller. If $f(x + k) = f(x)$ then $\sin(2x + 2k) = \sin 2x$ so the period of $\sin $ would be $2k$ or smaller. So $2k$ is not smaller than $2\pi$.)

Your confusion lies in you think we are adding $2\pi$ to $2x$ into the argument of $\sin$ it makes the period $2\pi$. True; it makes the period of SINE(x) $2\pi$. But we are sticking the $2\pi$ into $\sin ()$; we are not sticking it into $\sin (2x)$. We are only sticking $\pi$ into ...

... okay, look at this: $\sin (x)$ can be written as $\sin( [\backslash stick input here/])\backslash$ and $\sin(2x)$ can be written as $\sin(2\times [\backslash stick input here/])$. And $a + b$ can be written as $\text {a is the main thing} ---\text{ b is tacked on for the ride}$ or $a --tackon-- b$.

So $\sin( [\backslash 2x/])\backslash = \sin( [\backslash 2x -tackon- 2\pi/])\backslash $. So the period is $2\pi$

But $\sin(2x [\backslash x -tackon- \pi/])=\sin(2x [\backslash x/]--tackon-- 2\pi)$

$= \sin([\backslash 2\times x/]--tackon-- 2\pi)=$

$\sin([\backslash 2\times x --tackon-- 2\pi/])=$

$\sin ([\backslash 2\times x /]) = \sin(2\times [\backslash x /])$.

So the period of $\sin(2\times [\backslash put input here /]$ is $\pi$.

fleablood
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