Sin(x) = 0 iff x = kpi for some k in Z

[mattmns]

Well the title is basically the question.
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Show that sin(x) = 0 if and only if x/\pi is an integer.
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Some definitions.

Let z be a complex number, then:

\cos(z) = \frac{e^{iz} + e^{-iz}}{2}

\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}

We also have the power series definition, though I don't think it would be helpful here.

We have already proved the following, if they may be of use (most likely c,d, e):

(a) We have \sin^2(z) + \cos^2(z) = 1.

(b) We have \sin'(x) = \cos(x) and \cos'(x) = -\sin(x).

(c) We have \sin(-x) = -\sin(x) and \cos(-x) = \cos(x).

(d) We have \cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y) and \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)

(e) We have \sin(0) = 0 and \cos(0) = 1.

(f) We have e^{ix} = \cos(x) + i\sin(x) and e^{-ix} = \cos(x) - i\sin(x)

Also we proved in the previous part of this exercise that \cos(x+ \pi) = -\cos(x) and \sin(x+ \pi) = -\sin(x).

And we define \pi = \inf\{x\in (0,\infty): sin(x) = 0\}

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For the problem, I think the \Leftarrow is proved with a simple induction argument and using that \sin(-x) = -\sin(x). Though, if you have an interesting way to do it, please share

However, the \Rightarrow direction is seeming to give me some trouble. Any ideas here?

edit... I just got the wild idea of proving the contrapositive which seems like it would work well. If we write x= k\pi + r for some 0 < r <\pi. Then use the addition formula for sin to finally get sin(x) = 0 + sin(r). Which cannot be 0 since we know that \pi is the smallest positive number with sin(x) = 0. Sounds good to me. Always fun to come up with an idea after posting some huge mess

 

[Gib Z]

Heres an argument that may have also worked -

for any k in Z, k*pi can be written as pi + pi + pi +pi ...k times.

Every time we took away a pi, it would become -sin(x), then take away another pi, sin x again. Keep doing that until eventually there are no pi's left, - sin 0= sin 0 = 0.