As I was walking through campus today, I had an interesting question pop into my head: How can we prove that the period of $\tan(x)$ is $\pi$ rather than $2\pi$? The answer to this was extremely straightforward: We start off with $$\tan(x) = \tan(x + T) = {\tan(x) + \tan(T) \over 1 - \tan(x) \tan(T)}$$ to give us $$-\tan^2(x)\tan(T) = \tan(T)$$ $$0 = \tan(T) + \tan^2(x)\tan(T)$$ $$0 = \tan(T)[1 + \tan^2(x)]$$ $$\implies \tan(T) = 0\;\;\;\;\;\text{and}\;\;\;\;1 + \tan^2(x) = 0 \implies \text{No real solution for any $x\in\mathbb{R}$}$$ Which for $\tan(T) = 0 \implies {\sin(T) \over \cos(T)} = 0 \implies \sin(T) = 0$, we have $T = 0, \pi \implies T = \pi$ to show that the period of $\tan(x)$ is $\pi$ if we desire a nontrivial answer.

But I got stuck trying to do the same with $\sin(x)$. I tried:

$$\sin(x) = \sin(x + T) = \sin(x)\cos(T) + \sin(T)\cos(x)$$ $$\implies \sin(x)[1 - \cos(T)] = \sin(T)\cos(x)$$ $$\implies \tan(x) = {\sin(T) \over 1 - \cos(T)}$$

But I got stuck here. I'm not sure how to isolate a single trig function in terms of $T$.

I Googled this proof, but everyone either uses Taylor Expansions, Euler's Formula, or calculus. But I'm looking for an argument I could present to someone with knowledge of trigonometry and no more. Any ideas?

  • $\begingroup$ Hah. That's probably what's wrong. Thanks! $\endgroup$ – Decaf-Math May 12 '16 at 1:10
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    $\begingroup$ How are you defining $\sin(x)$ and $\cos(x)$? In your development of the periodicity of $\tan(x)$, you used knowledge of the tangent function at $\pi$ and presumably at all values less than $\pi$. So, it seems to be a circular argument. My suggestion is to start from a definition of the function. $\endgroup$ – Mark Viola May 12 '16 at 1:12
  • $\begingroup$ If $\sin (x+T)=\sin x$ for ALL $x$, and $\cos T \ne 0,$ then from your last line, $\sin x$ is a constant function. A similar argument, if $\tan T\ne 0 $,applies in the second equation regarding the period of $\tan x.$ $\endgroup$ – DanielWainfleet May 12 '16 at 2:13

If $T$ is such that for all $x\in \mathbb{R} $ we have $\sin(x+T) =\sin(x) $, then in particular, setting $x=0$, we have

$$\sin T =0$$

So $T=k\pi$ with $k \in \mathbb{Z} $. To conclude, we then need to check that $\sin(x+2\pi) =\sin(x)$ using your formula above (and that $\pi$ is not a period, by plugging $x=-\pi/2$ for example).

  • $\begingroup$ I forgot that this was possible: We're assuming that it holds for any $x\in\mathbb{R}$, so we're allowed to choose $x=0$ in particular and solve for $T$ that way. It's not the first thing in my mind because it feels too similar to proof by example which isn't valid. $\endgroup$ – Decaf-Math May 12 '16 at 1:33

Use the sum formulas $$\sin(x+T)=\sin x \cos T+\cos x\sin T$$ and the fact that $\sin (2\pi)=0$ and $\cos (2\pi)=1$ gives you a period of $2\pi$. Conversely is $\sin (x+T)=\sin x$ for all $x$ then $\sin T=0$ so $T=k\pi$ and one can easily see that $k$ must be even.


If $\sin(x+T)=\sin x,$ using Prosthaphaeresis Formula,

$$2\sin\dfrac T2\cos\left(x+\dfrac T2\right)=0$$ as $\cos\left(x+\dfrac T2\right)$ is dependent on $x,$

$\cos\left(x+\dfrac T2\right)=0$ won't give a constant value of $T$

So, we need $\sin\dfrac T2=0\iff\dfrac T2=n\pi$ where $n$ is any integer

$\implies T=?$


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