- #1
mattmns
- 1,128
- 6
Well the title is basically the question.
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Show that sin(x) = 0 if and only if [itex]x/\pi[/itex] is an integer.
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Some definitions.
Let z be a complex number, then:
[tex]\cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/tex]
[tex]\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/tex]
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 [itex]\sin^2(z) + \cos^2(z) = 1[/itex].
(b) We have [itex]\sin'(x) = \cos(x)[/itex] and [itex]\cos'(x) = -\sin(x)[/itex].
(c) We have [itex]\sin(-x) = -\sin(x)[/itex] and [itex]\cos(-x) = \cos(x)[/itex].
(d) We have [itex]\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)[/itex] and [itex]\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)[/itex]
(e) We have [itex]\sin(0) = 0[/itex] and [itex]\cos(0) = 1[/itex].
(f) We have [itex]e^{ix} = \cos(x) + i\sin(x)[/itex] and [itex]e^{-ix} = \cos(x) - i\sin(x)[/itex]
Also we proved in the previous part of this exercise that [itex]\cos(x+ \pi) = -\cos(x)[/itex] and [itex]\sin(x+ \pi) = -\sin(x)[/itex].
And we define [itex]\pi = \inf\{x\in (0,\infty): sin(x) = 0\}[/itex]
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For the problem, I think the [itex]\Leftarrow[/itex] is proved with a simple induction argument and using that [itex]\sin(-x) = -\sin(x)[/itex]. Though, if you have an interesting way to do it, please share
However, the [itex]\Rightarrow[/itex] 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 [itex]x= k\pi + r[/itex] for some [itex]0 < r <\pi[/itex]. Then use the addition formula for sin to finally get sin(x) = 0 + sin(r). Which cannot be 0 since we know that [itex]\pi[/itex] 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
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Show that sin(x) = 0 if and only if [itex]x/\pi[/itex] is an integer.
--------
Some definitions.
Let z be a complex number, then:
[tex]\cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/tex]
[tex]\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/tex]
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 [itex]\sin^2(z) + \cos^2(z) = 1[/itex].
(b) We have [itex]\sin'(x) = \cos(x)[/itex] and [itex]\cos'(x) = -\sin(x)[/itex].
(c) We have [itex]\sin(-x) = -\sin(x)[/itex] and [itex]\cos(-x) = \cos(x)[/itex].
(d) We have [itex]\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)[/itex] and [itex]\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)[/itex]
(e) We have [itex]\sin(0) = 0[/itex] and [itex]\cos(0) = 1[/itex].
(f) We have [itex]e^{ix} = \cos(x) + i\sin(x)[/itex] and [itex]e^{-ix} = \cos(x) - i\sin(x)[/itex]
Also we proved in the previous part of this exercise that [itex]\cos(x+ \pi) = -\cos(x)[/itex] and [itex]\sin(x+ \pi) = -\sin(x)[/itex].
And we define [itex]\pi = \inf\{x\in (0,\infty): sin(x) = 0\}[/itex]
---------
For the problem, I think the [itex]\Leftarrow[/itex] is proved with a simple induction argument and using that [itex]\sin(-x) = -\sin(x)[/itex]. Though, if you have an interesting way to do it, please share
However, the [itex]\Rightarrow[/itex] 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 [itex]x= k\pi + r[/itex] for some [itex]0 < r <\pi[/itex]. Then use the addition formula for sin to finally get sin(x) = 0 + sin(r). Which cannot be 0 since we know that [itex]\pi[/itex] 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
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