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