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.
$\endgroup$ 64 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $f(x)=\sin(2x)$: $$T_{\sin(2x)}=\inf_{t \in \mathbb R}\{t\mid \forall x\in \mathbb R:\sin(2x)=\sin(2x+2t)\}=\pi$$
$\endgroup$ $\begingroup$If $f(x)$ has period $p$, then $f(ax)$ has period $p/a$.
Proof
Let $g(x)=f(ax)$.
$g(x)=f(ax)=f(ax+p)=f(ax+a\cdot \frac{p}{a})=f(a(x+p/a))=g(x+p/a)$
$\endgroup$ $\begingroup$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 $\pi$.
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