A question on Cosine.

**Mathelogician** · June 22nd 2010, 11:16 PM

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

**undefined** · June 22nd 2010, 11:43 PM

Originally Posted by Mathelogician

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

Hi,

(Edited)

The full theorem is: $\cos x=\cos a \Rightarrow x=2k\pi\pm a$ for some integer k. In other words, $\cos x=\cos a \Rightarrow \exists k\in\mathbb{Z}<img src=$ x=2k\pi+a)\lor (x=2k\pi-a)" alt="\cos x=\cos a \Rightarrow \exists k\in\mathbb{Z}

x=2k\pi+a)\lor (x=2k\pi-a)" />. This is much different from $\cos x=\cos a \Rightarrow \forall k \in \mathbb{Z}:x=2k\pi\pm a$ . Among other things, this would allow us to easily show that $0=2\pi$ .

Regarding writing math formulae, etc., you can see the LaTeX tutorial within the LaTeX Help Subforum.

**Mathelogician** · June 23rd 2010, 12:21 AM

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!
---------

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

**undefined** · June 23rd 2010, 12:34 AM

Originally Posted by Mathelogician

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!
---------

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

By you argument, $cos(0)=cos(2\pi)\Rightarrow \forall k\in\mathbb{Z}<img src=$ 0=2k\pi+2\pi)\lor(0=2k\pi-2\pi)" alt="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 $(0=4\pi+2\pi=6\pi)\lor(0=4\pi-2\pi=2\pi)$ . This is a contradiction.

What exactly is your question?

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

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

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

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

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. $\forall a,b\in\mathbb{Z},|a|+|b|\ne0:a|b\Leftrightarrow \exists k\in\mathbb{Z}:ka=b$ .

**Mathelogician** · June 23rd 2010, 04:43 AM

Originally Posted by undefined

By you argument, $cos(0)=cos(2\pi)\Rightarrow \forall k\in\mathbb{Z}<img src=$ 0=2k\pi+2\pi)\lor(0=2k\pi-2\pi)" alt="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 $(0=4\pi+2\pi=6\pi)\lor(0=4\pi-2\pi=2\pi)$ . This is a contradiction.

What exactly is your question?

Edit: Actually, I should have written that the full theorem is:
$\forall x,a\in\mathbb{R}:\cos x=\cos a \Rightarrow x=2k\pi\pm a$ for some integer k.

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: $\forall x,a\in\mathbb{R}, k\in\mathbb{Z}:\cos x=\cos a \Leftrightarrow x=2k\pi\pm a$ . Infact we have infinite solutions for that, and for k=0 we have 2 solutions like other amounts of k.

**Defunkt** · June 23rd 2010, 05:09 AM

Originally Posted by Mathelogician

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: $\forall x,a\in\mathbb{R}, k\in\mathbb{Z}:\cos x=\cos a \Leftrightarrow x=2k\pi\pm a$ . Infact we have infinite solutions for that, and for k=0 we have 2 solutions like other amounts of k.

When you say $cosx = cosa$ has solutions $x = 2 \pi k \pm a$ , what do you mean?
You mean that there exists such a $k \in \mathbb{N}$ , for which $x = 2 \pi k + a$ . It does not mean that $x = 2 \pi k \pm a$ for all $k \in \mathbb{N}$ !

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

What does this mean? It means that if you have a number, say $a$ , for which $a^2 - a - 2 = 0$ , then either $a = 2$ or $a = -1$ . It does not mean that $a = 2 = -1$ !
The case for $cosx = cosa$ is exactly the same!

**undefined** · June 23rd 2010, 05:20 AM

Originally Posted by Mathelogician

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: $\forall x,a\in\mathbb{R}, k\in\mathbb{Z}:\cos x=\cos a \Leftrightarrow x=2k\pi\pm a$ . Infact we have infinite solutions for that, and for k=0 we have 2 solutions like other amounts of k.

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.

**Ackbeet** · June 23rd 2010, 05:49 AM

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.

**p0oint** · June 23rd 2010, 07:07 AM

Ok, Why you make the things complicated than they are?

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

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

Now this works for every x in reals.

**Ackbeet** · June 23rd 2010, 07:15 AM

Mathelogician: what are you trying to do? I'm just taking a step back here. What is your goal?

**Mathelogician** · June 23rd 2010, 08:38 AM

Hello and Thanks for responses.
------------------------------
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!

**Ackbeet** · June 23rd 2010, 09:03 AM

I would agree with your claim, mathelogician. It is true that the sin and cos functions are $2\pi$ -periodic. So, $\cos(x)=\cos(x+2\pi k)\;\forall\,k\in\mathbb{Z}$ , and $\forall\,x\in\mathbb{R}$ . However, reasoning backwards to any sort of equality of the x's is incorrect. Example:

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

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

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 $\cos(x)=\cos(a)$ THEN $x=2\pi k\pm a$ . But I've just shown you a counterexample to that claim. The converse of that claim, that IF $x=2\pi k\pm a$ THEN $\cos(x)=\cos(a)$ , 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.

**Mathelogician** · June 23rd 2010, 11:53 AM

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).

--------

Why do you insist on asking my goal of questioning??!!!

**Ackbeet** · June 23rd 2010, 12:11 PM

Your Original Claim: if $\cos(x)=\cos(a)$ , then $x=2\pi k\pm a$ for all k. I would love to see a proof of this false claim. Here is my proof that it is false:

Let $a=\pi/3$ . Then $\cos(-\pi/3)=\cos(\pi/3)$ , and yet it is not true that $-\pi/3=2\pi k\pm\pi/3$ for any integer $k$ . Therefore, the implication in the claim is false.

**undefined** · June 23rd 2010, 12:56 PM

Originally Posted by Mathelogician

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.

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.

Originally Posted by Mathelogician

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

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 $\displaystyle \Leftrightarrow$ 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.

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