Prove Sin(x) Unique for Every Positive Integer

[cragar]

Homework Statement

Prove that the sin(x), where x is a positive integer. For every value of x has a distinct unique number.
For example sin(3)≠sin(5)

The Attempt at a Solution

I thought about using the infinite series for sin(x), or Euler's formula , Or maybe since an integer doesn't divide into pi so therefore it won't repeat the cycle.

 

[Curious3141]

Homework Statement

Prove that the sin(x), where x is a positive integer. For every value of x has a distinct unique number.
For example sin(3)≠sin(5)

The Attempt at a Solution

I thought about using the infinite series for sin(x), or Euler's formula , Or maybe since an integer doesn't divide into pi so therefore it won't repeat the cycle.

Try a proof by contradiction. Can you think about what \sin x = \sin y would imply about the relationship between x and y? Now let x and y represent distinct positive integers and arrive at a contradiction.

 

[cragar]

thanks for the help, if we assumed sin(x)=sin(y) and they were positive integers, we would need one of them to be an integer multiple of pi, because we would need another value to eventually come back around to one of the other values. But this will never happen because we are dealing with integers.

 

[Curious3141]

thanks for the help, if we assumed sin(x)=sin(y) and they were positive integers, we would need one of them to be an integer multiple of pi, because we would need another value to eventually come back around to one of the other values. But this will never happen because we are dealing with integers.

It's the difference between them that has to be an integer multiple of 2\pi.

I was thinking along the lines of:

\sin x = \sin y \Rightarrow x = y + 2n{\pi}, n \in \mathbb{Z}

Without loss of generality we may assume x > y:

(x > y) \vee (x,y \in \mathbb{Z^+}) \Rightarrow n > 0

Hence,

\pi = \frac{x-y}{2n} \Rightarrow \pi \in \mathbb{Q}

Contradiction.

 

[Dick]

It's the difference between them that has to be an integer multiple of 2\pi.

I was thinking along the lines of:

\sin x = \sin y \Rightarrow x = y + 2n{\pi}, n \in \mathbb{Z}

Without loss of generality we may assume x > y:

(x > y) \vee (x,y \in \mathbb{Z^+}) \Rightarrow n > 0

Hence,

\pi = \frac{x-y}{2n} \Rightarrow \pi \in \mathbb{Q}

Contradiction.

sin(x)=sin(y) does not imply x=y+2npi. sin(pi/2-1/2)=sin(pi/2+1/2). The arguments of the sin don't differ by a multiple of 2pi.

 

[Curious3141]

sin(x)=sin(y) does not imply x=y+2npi. sin(pi/2-1/2)=sin(pi/2+1/2). The arguments of the sin don't differ by a multiple of 2pi.

Oops. Thanks for spotting that!

The first line of the proof can be easily amended to:

\sin x = \sin y \Rightarrow x = y + 2n{\pi}, n \in \mathbb{Z} OR x+y = (2k+1)\pi, k \in \mathbb{Z}

With the same conclusion that \pi is rational following from either case.