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Can it be proven that

$ \forall x, y \in \mathbb{Z} \left( \sin(x) = \sin(y) \iff x = y\right)$

? Or disproven, of course. And likewise with cosine?

Since sine and cosine have periods of $2\pi$, for $x$ and $y$ radians to have the same location on the unit circle, $x$ and $y$ would have to differ by $2k\pi$ radians, where $k \in \mathbb{Z}$.

How can I show that there exists no $k$ such that $y - 2k\pi = x$, given $x, y, k \in \mathbb{Z}$?

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It's not true that if $\sin(x)=\sin(y)$ then $x$ and $y$ differ by $2k\pi$ radians. –  Chris Eagle Jul 21 '13 at 16:28
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It comes down to proving that $\pi$ is irrational, or, more likely, quoting the result. What you need is an easy consequence. –  André Nicolas Jul 21 '13 at 16:29
    
Oh yeah, you're right. I forgot about $y = \pi - x$ and the similar, negative quadrant version. –  scott_fakename Jul 21 '13 at 16:30

2 Answers 2

up vote 2 down vote accepted

You are correct. It should be clear why $\forall x,y\in\Bbb{Z}$, $x=y\implies\sin(x)=\sin(y)$. It remains to prove the converse: $\forall x,y\in\Bbb{Z}$, $\sin(x)=\sin(y)\implies x=y$

We proceed by contradiction. Suppose instead that there exist integers $x,y$ such that $\sin(x)=\sin(y)$ yet $x\neq y$. Then, as you figured out, there are only $2$ possibilities:

Case 1: Suppose that $y=x+(2\pi)k$ for some nonzero $k\in \Bbb{Z}$. Then solving for $\pi$, we obtain: $$ \pi=\dfrac{y-x}{2k} $$ Since $x,y,k\in\Bbb{Z}$ where $k\neq0$, it follows that $\pi \in \Bbb{Q}$, a contradiction.

Case 2: Suppose that $y=(\pi-x)+(2\pi)k$ for some $k\in \Bbb{Z}$. Then solving for $\pi$, we obtain: $$ \pi=\dfrac{y+x}{1+2k} $$ Since $x,y,k\in\Bbb{Z}$ where $k\neq-1/2$, it follows that $\pi \in \Bbb{Q}$, a contradiction.

In either case, we obtained a contradiction. Hence, we have proven the converse, as desired.

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This is equivalent to $\pi$ being irrational. So it can be proven, yes.

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