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Proving a trig identity: set theoretic ideas? 
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#1
Jan809, 06:11 AM

P: 76

Hi everyone,
Any assistance with this following problem would be greatly appreciated. I'm in Year 11 and working through Apostol volume 1. 1. The problem statement, all variables and given/known data sin n*pi = 0, where n is an integer sin n*pi =/= 0, where n is not an integer. Prove these statements... 2. Relevant equations Hmm... 3. The attempt at a solution My idea is that perhaps I need to use set theoretic ideas or mathematical induction, initially to show that the first case holds for all n. sin 0*pi = 0 sin 1*pi = 0 sin (1+1)pi = 0 = sin (n+1)*pi sin n*pi = 0 So, sin (n+1)*pi = 0 (where n is an integer) Now, to use set theoretic notation: Let sin n*pi = A. A = {n is an element of Rsin n*pi = 0, n is an integer} B = {n is an element of Rsin n*pi =/= 0, n is not an integer} .: A = {0} .: B =/= {0} Now, this is circular  I haven't proved anything at all. (sorry for not using latex, unsure about how to write set notation). Any assistance would be grand. Cheers, Davin 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Jan809, 06:51 AM

HW Helper
P: 3,352

The general layout of proof is to first show that an initial case is true, then to assume it is true for some case n=k. This assumption is called the induction hypothesis. The final step is to consider the case n=k+1, and using the hypothesis, showing that if n=k holds, then so must n=k+1. Then we remember n=0 holds, so n=1 holds, so n=2 holds... and etc it holds for all integers.
So in this case, first step is done, [itex]sin (0*\pi) = 0[/itex]. You did this step correctly. Next step, we assume that, for some k, the statement [itex] sin (k\pi) = 0[/itex] holds true. Now you must consider the next case, where n=k+1. You have to use your induction hypothesis to show the statement for n=k+1 is true as well. In order to use your induction hypothesis, use the expansion formula for sin (x+y). You should be able to finish the rest off easily. Just a side note that the set notation you used isn't expressed correctly, and in any case don't help your cause. Generally when working out of a textbook, a good way to go is to use similar methods to the author. For the next part, keeping in mind what we just proved, contradiction sounds a good way to go, good luck! 


#3
Jan909, 06:04 AM

P: 76

Hi Gib Z,
This is not so clearcut  you state that I would be able to obtain the results easily using the expansion property: not so. I get sin(x+y) = sin x cos y + cos x sin y, however do not forget I am working with the property sin(x+y)*pi... Perhaps I need to multiply the result by sin pi but I am deeply aware that this can not be the case, as this is equal to zero. This would counter the argument (incorrectly) when the variable x is not an integer. This is my only formal treatment of trigonometric functions so far, so what I lead myself into thinking is purely from my own understanding from the text. Perhaps I need to solve for sin(x*pi + y), and no doubt this equals sin x*pi cos y + cos x*pi sin y (which I believe I have confirmed, but not proven, through experiment, at least when y = 0). My real weakness in terms of mathematics at this stage is proofs  mostly I do not have too many issues with anything else. Any assistance is valuable  I just do not see the logic in the idea of proofs, I'm afraid. The real reason behind this, no doubt, is that I've done so little proof writing (except copying down and reading Apostol's proofs  which admittedly I can understand). Many people say that reading proofs is the best way to learn how to write them: so why doesn't this sink in?? I've read numerous proofs! The only proofs I can work generally are those based on mathematical induction (not in this case as a trig identity though). Cheers, Davin 


#4
Jan909, 06:15 AM

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Proving a trig identity: set theoretic ideas?



#5
Jan909, 07:48 PM

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The best way to learn to write proofs is to practice writing proofs! Like any fine art, you'll start off struggling, sloppy, and overall not very good. Practice makes perfect. 


#6
Jan909, 07:54 PM

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The trick here is to build upon your previous knowledge. You know a lot about the sin function. In particular, you know that all of its relevant behavior is demonstrated on the interval [itex][0, \pi/2][/itex], and you know how to relate that interval to any other place on the real line.
(In other words, you know the symmetries of the sine function  e.g. it's periodic and odd) Use that knowledge to solve this problem. Sketch an outline of the proof before you try to actually write the proof. What ideas are you going to use? What is the precise statement of those ideas? 


#7
Jan909, 09:37 PM

P: 76

Hi Hurkyl,
Yes, I recognise the symmetries of the sine function. I can show that sin n*pi is equal to zero, when n is an integer simply by drawing a diagram (which I have done). I am also well aware that sin n*pi does not equal zero when n is not an integer, intuitively. However, I am unsure of how to bridge the gap between showing and proving. In each case, I should be able to use sin (k+1)pi with n = k+1 (the first with k an integer, the second with k not an integer). Perhaps in the second case, I need to use a small, nonzero value such as 0.001 and show that for this case, sin n*pi is not equal to zero. I would then generalise and use the induction hypothesis, with k+1. Also, Gib Z, as a side note I agree with you regarding practice writing proofs. So it really is just like learning an art then. It happens to be that I am a good writer, but I don't enjoy it much any more. However, that's beside the point! I do see the point you are making though. Cheers, Davin 


#8
Jan1209, 03:15 AM

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P: 2,263

When solving these types of problems you need to be aware of your assumptions. Are you only considering real numbers? What is you definition of sine? What is you definition of pi? What relevant theorems do you have?
Your question can be rephrased find all real x such that sin(x)=0 A possible definition of pi is let pi be the unique (strictly) positive real number such that whenever 0<x<pi sin(x)>0 let us use sin(x)=sin(x) so that we may consider without loss of generality only positive x sin(0)=0 sin(x+pi)=sin(x) so sin(0)=0=>sin(pi)=0=>sin(2pi)=0=>... by induction sin(n*pi)=0 n any integer likewise (0<x<pi=>sin(x)>0)=>(pi<x<2pi=>sin(x)>0)=>(2pi<x<3pi=>sin(x)>0)=> ... by induction for any integer n (n1)pi<x<n*pi=>sin(x)>0 


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