# Why is the period of sin(2x) π?

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.

• 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 3 '17 at 18:05
• 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 3 '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 3 '17 at 18:07
• This boils down to: If an object spins twice as fast, it takes half as long to make one turn. – Semiclassical May 3 '17 at 18:08
• @Semiclassical, I can understand this intuitively but it is the reasoning which is puzzling me. – MrAP May 3 '17 at 18:10

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 $\pi$.

• You may very well wanna make an edit don't u :). – Megadeth May 3 '17 at 18:55

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.

• How "by definition we proved that $x↦sin(2x)$ has period = $π$"? Are you referring to the definition of a periodic function? – MrAP May 3 '17 at 18:38
• Indeed. I assumed you were given the most commonly seen definition. To avoid unnecessary misunderstandings I amended the sentence. – Megadeth May 3 '17 at 18:54

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 $f(x)=\sin(2x)$: $$T_{\sin(2x)}=\inf_{t \in \mathbb R}\{t\mid \forall x\in \mathbb R:\sin(2x)=\sin(2x+2t)\}=\pi$$