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Math Help - A question on Cosine.

  1. #16
    Member Mathelogician's Avatar
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    Dear Ackbeet and undefined,
    I bet you still have not understood my assertion!
    I told you the theorem states for Trig EQUATIONS. It means that for trig equations we must have an unknown value called x and obtain all of the possible values for that.This is the meaning of Trigonometric equation(similar to the meaning of a linear equation or a quadratic or... wich we look for unknown x) and the solution set is nothing except that!
    And you say is called a Trig IDENTITY.
    --------------------------------------------------------------
    You see in the following picture a Geometric proof from the book مثلثات (Trigonometry) by Ahmad Ghandehary Madreseh publications:

    You can see another proof for that here:
    Trigonometry - Matzic
    And can see its uses in many books and Articles; for example:
    Plane Trigonometry, by S.L LONEY– Cambridge University Press.
    Elementary Trigonometry , by H.S Hall and S. R. KNIGHT
    مثلثات (Trigonometry) by Ahmad Ghandehary - Madreseh publications

    Even you can google the word “Trigonometric Equations” and find useful articles and books for that!
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  2. #17
    A Plied Mathematician
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    My posts 12 and 14 are incorrect, on second look, and looking at your proof. Please disregard those two posts, and accept my apologies for being so thick-headed.

    Just looking back at the original post: are you claiming that the equation cos(x) = cos(a) implies x = 0?
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  3. #18
    Member Mathelogician's Avatar
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    Your welcome.
    I think your question is:
    cos(x) = cos(0) implies x = 0 ?
    If so, I should say that: It implies x=2k*pi and x=0 is one of the infinte solutions which is obtained by getting k=0 and infact it is the original angle.
    And if no, cos(x) = cos(a) implies x=2k*pi+a and 2k*pi-a.
    Last edited by Mathelogician; June 24th 2010 at 09:47 PM. Reason: Bad typing!
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  4. #19
    MHF Contributor undefined's Avatar
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    In post #15, I wrote that I would only post again in this thread to correct any mistakes I had made; this was on grounds that I felt the thread to be unproductive; I now feel this thread has taken a productive turn, so I am posting again, even though I do not see a mistake in my previous posts. (Not to say I haven't made any mistake, just that I don't see one.)

    I think the main problem in this thread is lack of clarity. I still have no idea what Mathelogician's question is, or if Mathelogician even has a question.

    Quote Originally Posted by Mathelogician View Post
    Dear Ackbeet and undefined,
    I bet you still have not understood my assertion!
    The theorem whose proof you presented is no different from

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

    Equivalently:

    \exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="\forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow x\in\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" />.

    The \forall quantifier in the proof you posted means that k ranges over all integers. The proof you posted does not say that x=2k\pi\pm\theta,\forall k\in\mathbb{Z}. This is a somewhat subtle point I suppose; the placement of the \forall quantifier is very important. Consider the two statements:

    (1) "The complete solution set of x is obtained by 2k\pi\pm\theta,\forall k\in\mathbb{Z}."

    (2) " x=2k\pi\pm\theta,\forall k\in\mathbb{Z}."

    Statement (1) says that x belongs to the set obtained by ranging over all integers, while statement (2) says the equation involving x holds for all integers. (And the equation does not hold for all integers, so statement (2) is false.)

    To translate explicitly: Statement (1) can be written:

    x\in\{0\cdot2\pi+\theta,0\cdot2\pi-\theta,1\cdot2\pi+\theta,1\cdot2\pi-\theta,\cdots \}

    And statement (2) can be written:

    (x=0\cdot2\pi+\theta\lor x=0\cdot2\pi+\theta) \land (x=1\cdot2\pi+\theta\lor x=1\cdot2\pi+\theta) \land \cdots

    The problem is those \land operators, which ought to be \displaystyle\lor instead. To accomplish this, we can use \exists.

    I'm sure there's a bit of a language barrier too, in terms of English. Anyway I hope you (Mathelogician) can see that you and I are talking about the exact same theorem, just with different language (and disagreement over the language). However, if you look carefully at everything I wrote, I believe you will find it's all meticulously in keeping with conventions used in math textbooks and journals and encyclopedias, etc. If not, then I'll be glad if someone will correct me.

    So, I understood your assertion, it's just that the assertion you wrote down was different from the assertion you meant to write down.

    Now.. Is there a question we can help answer?
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  5. #20
    Member Mathelogician's Avatar
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    Lightbulb

    NO;
    1-My claim (using Universal quantifier) is different from yours(using existensial quantifier)!
    When you say: There exists a number k in Z such that:...", it means that "for SOME (at least one) k in Z,we have ..."; But it doesn't clear our mean which is "for EVERY k in Z we have..."
    Indeed yours is true but inadequate and doesn't make the real sense!
    2-You need to know that the expression "x=2k*pi+_ a for all k in Z" is ANOTHER FORM of the first expression which is used in some books and articles; And since you don't know this, you insist on your own interpretation of that and think they are different!!
    I thought you know this and then i used it for the proof!
    NOTE: Existential quantification is distinct from Universal quantification ("for all"), which asserts that the property or relation holds for any members of the domain.
    See:
    Existential quantification - Wikipedia, the free encyclopedia
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  6. #21
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Mathelogician View Post
    NO;
    1-My claim (using Universal quantifier) is different from yours(using existensial quantifier)!
    When you say: There exists a number k in Z such that:...", it means that "for SOME (at least one) k in Z,we have ..."; But it doesn't clear our mean which is "for EVERY k in Z we have..."
    Indeed yours is true but inadequate and doesn't make the real sense!
    2-You need to know that the expression "x=2k*pi+_ a for all k in Z" is ANOTHER FORM of the first expression which is used in some books and articles; And since you don't know this, you insist on your own interpretation of that and think they are different!!
    I thought you know this and then i used it for the proof!
    NOTE: Existential quantification is distinct from Universal quantification ("for all"), which asserts that the property or relation holds for any members of the domain.
    See:
    Existential quantification - Wikipedia, the free encyclopedia
    The set \exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" /> equals the set \{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots  \}.

    Thus

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

    is equivalent to

    \forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow x\in\{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots   \}

    which says that \{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots   \} is the complete solution set of \displaystyle x, which is your theorem.

    ---------------------------

    Let's demonstrate that the set \exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" /> equals the set \{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots   \}.

    Consider 0\cdot2\pi+a. Does there exist a k\in\mathbb{Z} such that 0\cdot2\pi+a=2k\pi\pma? Yes, so it's in the set. ( k=0.)

    Consider 0\cdot2\pi-a. Does there exist a k\in\mathbb{Z} such that 0\cdot2\pi-a=2k\pi\pma? Yes, so it's in the set. ( k=0.)

    Consider 1\cdot2\pi+a. Does there exist a k\in\mathbb{Z} such that 1\cdot2\pi+a=2k\pi\pma? Yes, so it's in the set. ( k=1.)

    etc.

    ------------------------------

    Let's demonstrate that

    \exists  k\in\mathbb{Z}: x=2k\pi\pm a

    is equivalent to

    \exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="x\in\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" />.

    Consider 0\cdot2\pi+a. Does there exist a k\in\mathbb{Z} such that 0\cdot2\pi+a=2k\pi\pma? Yes, so it satisfies the first statement, and the second.

    Etc.

    ---------------------------

    If you've studied equivalence classes then you will recognise that this is what we are talking about. cos x = cos a if and only if x and a are the in the same equivalence class defined by the condition \exists  k\in\mathbb{Z}: x=2k\pi\pm a.
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  7. #22
    Member Mathelogician's Avatar
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    Thumbs up

    Ok,
    You are right.
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