Hi all,

I did'nt know how to write formula here. So:

http://i47.tinypic.com/2hg4e52.jpg

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- Jun 22nd 2010, 11:16 PMMathelogicianA question on Cosine.
Hi all,

I did'nt know how to write formula here. So:

http://i47.tinypic.com/2hg4e52.jpg - Jun 22nd 2010, 11:43 PMundefined
Hi,

(Edited)

The full theorem is: $\displaystyle \cos x=\cos a \Rightarrow x=2k\pi\pm a$ for some integer k. In other words, $\displaystyle \cos x=\cos a \Rightarrow \exists k\in\mathbb{Z}:(x=2k\pi+a)\lor (x=2k\pi-a)$. This is much different from $\displaystyle \cos x=\cos a \Rightarrow \forall k \in \mathbb{Z}:x=2k\pi\pm a$. Among other things, this would allow us to easily show that $\displaystyle 0=2\pi$.

Regarding writing math formulae, etc., you can see the LaTeX tutorial within the LaTeX Help Subforum. - Jun 23rd 2010, 12:21 AMMathelogician
Thanks.

No my assert is right! We know that k is an element of the set of integers and that solution is the general solution of the equation and it means that for all ks it is true!

So if we choose k to be 0, then x=+-x and we choose the case x=-x wich is one of the solutions!

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Even If that is wrong, when we get x=2k(pi)+-x the there are 2 cases: 1) 2x=2k*pi=>x=k*pi and 2)**2k*pi=0 => k=0 wich we may have choiced k=another number except 0**for example if k=1 =>**2*pi=0** - Jun 23rd 2010, 12:34 AMundefined
By you argument, $\displaystyle cos(0)=cos(2\pi)\Rightarrow \forall k\in\mathbb{Z}:(0=2k\pi+2\pi)\lor(0=2k\pi-2\pi)$, and letting k=2, we have $\displaystyle (0=4\pi+2\pi=6\pi)\lor(0=4\pi-2\pi=2\pi)$. This is a contradiction.

What exactly is your question?

Edit: Actually, I should have written that the full theorem is:

$\displaystyle \forall x,a\in\mathbb{R}:\cos x=\cos a \Rightarrow x=2k\pi\pm a$ for some integer k.

Furthermore, the converse is true. So, for the most information and least amount of characters:

$\displaystyle \forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow\exists k\in\mathbb{Z}: x=2k\pi\pm a$.

Edit 2: I suppose it's worth mentioning that for the correct theorem, we can't just choose an arbitrary k, like you did at the end the quoted post.

Maybe it would help you to have an analogy. $\displaystyle \forall a,b\in\mathbb{Z},|a|+|b|\ne0:a|b\Leftrightarrow \exists k\in\mathbb{Z}:ka=b$. - Jun 23rd 2010, 04:43 AMMathelogician
1) I see this is a contradiction and i want to know why it hapens when we use allowable ways!!

2) And the theorem is:$\displaystyle \forall x,a\in\mathbb{R}, k\in\mathbb{Z}:\cos x=\cos a \Leftrightarrow x=2k\pi\pm a$. Infact we have**infinite**solutions for that, and for k=0 we have 2 solutions like other amounts of k. - Jun 23rd 2010, 05:09 AMDefunkt
When you say $\displaystyle cosx = cosa$ has solutions $\displaystyle x = 2 \pi k \pm a$, what do you mean?

You mean that**there exists**such a $\displaystyle k \in \mathbb{N}$, for which $\displaystyle x = 2 \pi k + a$. It**does not**mean that $\displaystyle x = 2 \pi k \pm a$**for all**$\displaystyle k \in \mathbb{N}$!

Take, for example, a quadratic equation - say $\displaystyle x^2 - x - 2 = 0$. We know, then, its solutions are $\displaystyle x_{1, 2} = \frac{1 \pm \sqrt{1 + 8}}{2} \Rightarrow x_1 = 2, \ x_2 = -1$

What does this mean? It means that if you have a number, say $\displaystyle a$, for which $\displaystyle a^2 - a - 2 = 0$, then**either**$\displaystyle a = 2$**or**$\displaystyle a = -1$. It does not mean that $\displaystyle a = 2 = -1$!

The case for $\displaystyle cosx = cosa$ is exactly the same! - Jun 23rd 2010, 05:20 AMundefined
1) You've heard of proof by contradiction, right? What I gave was a proof that your claim is false. You have not used "allowable ways."

2) This is false, as was already proven.

In general: You are using "proof by vehement assertion." You are not presenting an actual proof, you are just stating emphatically that your claim is true, over and over.

See Defunkt's post which goes along with everything I've been saying. - Jun 23rd 2010, 05:49 AMAckbeet
Indeed, the location of quantifiers in logic ("there exists", or, "for all") is extremely important, as demonstrated above. If a statement looks the same, but has those quantifiers in different locations, then it's

*not necessarily the same statement*. - Jun 23rd 2010, 07:07 AMp0oint
Ok, Why you make the things complicated than they are?

$\displaystyle x = 2k\pi \pm a$ for cos(a)=cos(x).

cos(-x)=cos(x) because cosine is even function.

Now this works for every x in reals. - Jun 23rd 2010, 07:15 AMAckbeet
Mathelogician: what are you trying to do? I'm just taking a step back here. What is your goal?

- Jun 23rd 2010, 08:38 AMMathelogician
Hello and Thanks for responses.

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Dear Defunkt and other friends, i mean that it's true for**all the integer numbers**.

For example: cos(x)=cos(2*pi+x)=cos(4*pi+x)=cos(6*pi+x)=cos(8*p i+x)=cos(10*pi+x)=cos(12*pi+x)=cos(14*pi+x)=cos(16 *pi+x)=cos(18*pi+x)=cos(20*pi+x)=cos(22*pi+x)=cos( 24*pi+x)=cos(26*pi+x)=cos(28*pi+x)=cos(30*pi+x)=.. ..{also for negative numbers}

If you use the Unit circle you will understant my assertion. Infact the Original period of Cosine function is p=2*pi (like sine function). If a function f**is periodic**with period P, then for all x in the domain of f and**all integers n**, f(x + nP) = f(x).

See: Periodic function - Wikipedia, the free encyclopedia

And the quadratic equation like any other polynomial**IS NOT**periodic.

So i think my assertion is Reasonable and yours is not! - Jun 23rd 2010, 09:03 AMAckbeet
I would agree with your claim, mathelogician. It is true that the sin and cos functions are $\displaystyle 2\pi$-periodic. So, $\displaystyle \cos(x)=\cos(x+2\pi k)\;\forall\,k\in\mathbb{Z}$, and $\displaystyle \forall\,x\in\mathbb{R}$. However, reasoning backwards to any sort of equality of the x's is incorrect. Example:

$\displaystyle \frac{1}{2}=\cos\left(\frac{\pi}{3}\right)=\cos\le ft(-\frac{\pi}{3}\right)$.

But, obviously, $\displaystyle \frac{\pi}{3}-\left(-\frac{\pi}{3}\right)=\frac{2\pi}{3}\not=2\pi k$ for any integer $\displaystyle k$.

The cosine and the sine functions are not 1-1; hence, reasoning from equality of the functions to equality of the arguments is not permissible. Reasoning from equality of arguments to equality of functions is permissible, since the cosine and sine functions are well-defined.

So, mathelogician, I would say that your original claim in the OP is incorrect. Your original claim was that IF $\displaystyle \cos(x)=\cos(a)$ THEN $\displaystyle x=2\pi k\pm a$. But I've just shown you a counterexample to that claim. The converse of that claim, that IF $\displaystyle x=2\pi k\pm a$ THEN $\displaystyle \cos(x)=\cos(a)$, is true. A statement is not, in general, equal to its converse!

But, all of this could well be irrelevant. I'm still left wondering what it is you're trying to do. - Jun 23rd 2010, 11:53 AMMathelogician
Thanks.

I think i got the Mistake!

1) When we speak about an equation, then**we must have an unknown number**(called x) an**we want to find all Possible values for x**. So my mistake was forgetting this Important issue!!

2) Then you should note that my claim wich is the expression and its converse, is ALWAYS true for Trig*EQUATIONS*.

Infact there we have an unknown number x and the General solution is the set of all possible values for that!

There are different proofs for this claim. For example a Geometric proof exists for that(If you need, i will write it here).

And almost in every Trigonometry book in you can find this solving method.(tell me if you need).

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Why do you insist on asking my goal of questioning??!!! - Jun 23rd 2010, 12:11 PMAckbeet
Your Original Claim: if $\displaystyle \cos(x)=\cos(a)$, then $\displaystyle x=2\pi k\pm a$ for all k. I would love to see a proof of this false claim. Here is my proof that it is false:

Let $\displaystyle a=\pi/3$. Then $\displaystyle \cos(-\pi/3)=\cos(\pi/3)$, and yet it is not true that $\displaystyle -\pi/3=2\pi k\pm\pi/3$ for any integer $\displaystyle k$. Therefore, the implication in the claim is false. - Jun 23rd 2010, 12:56 PMundefined
I don't know where you get your definitions. 5=5 is an equation. And clearly in the equation 5=5 there is no unknown.

Just to be perfectly clear that we are using the same language: for a statement

p -> q

the converse is

q -> p

Ackbeet very clearly explained that using "for all integers k" only one direction is true, while the other is false. Please try to understand this. Using the symbol $\displaystyle \displaystyle \Leftrightarrow$ is to claim that both directions are true.

I will not post on this thread anymore (except to correct any mistake I may have made) because I feel the discussion is becoming unproductive, just saying the same thing over and over. Hope all this makes sense to the OP. Cheers.