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Math Help - [SOLVED] Trig Functions

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question [SOLVED] Trig Functions

    This Is Really Bad, Ive Frgot What To Do With The Problem Below:

    y= 3csc (3x+pi) - 2

    it asks for :
    1. the period
    2. domain
    3. range
    4 the graph...but there is like something that helps to clue you in graphing, i dont remember...im reading through the book.

    this is what i did with the equation:

    3csc [3(x + pi/3)] - 2
    where do i go after that?
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  2. #2
    Moo
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    Buenos dias
    Quote Originally Posted by >_<SHY_GUY>_< View Post
    This Is Really Bad, Ive Frgot What To Do With The Problem Below:

    y= 3csc (3x+pi) - 2

    it asks for :
    1. the period
    2. domain
    3. range
    4 the graph...but there is like something that helps to clue you in graphing, i dont remember...im reading through the book.

    this is what i did with the equation:

    3csc [3(x + pi/3)] - 2
    where do i go after that?
    For the period, let t be the period, t>0.
    f(x+t)=f(x)
    f(x+t)=3 csc (3x+3t+pi)-2
    but you know that csc has a period 2pi, that is to say csc(z+2pi)=csc(z)
    so 3t=2pi
    --> t=2pi/3

    the domain is the set of all real numbers, except the ones for which csc is undefined (which are the ones that annulate a sine) :
    solve for x in 3x+pi=k*pi, for any (positive or negative) integer k.


    do these help ?
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by Moo View Post
    Buenos dias

    For the period, let t be the period, t>0.
    f(x+t)=f(x)
    f(x+t)=3 csc (3x+3t+pi)-2
    but you know that csc has a period 2pi, that is to say csc(z+2pi)=csc(z)
    so 3t=2pi
    --> t=2pi/3

    the domain is the set of all real numbers, except the ones for which csc is undefined (which are the ones that annulate a sine) :
    solve for x in 3x+pi=k*pi, for any (positive or negative) integer k.


    do these help ?
    Ah! Bonour Mademoiselle "Moo" =]
    Comment Allez-Vous?

    ok, i got how to determine the period....Muchas Gracias


    but the only thing that really confuses me is finding domain and ranges of graph, especially for trig functions :/
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  4. #4
    Moo
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    Ah! Bonour Mademoiselle "Moo" =]
    Comment Allez-Vous?

    ok, i got how to determine the period....Muchas Gracias


    but the only thing that really confuses me is finding domain and ranges of graph, especially for trig functions :/
    ša va, merci

    \csc(z)=\frac{1}{\sin(z)}
    So it's not defined if \sin(z)=0.
    \sin(z)=0 \Leftrightarrow z=k \pi ~,~ k \in \mathbb{Z} (the set of all integers)

    So for \csc(3x+\pi) to be defined, you need 3x+\pi to be different from k \pi
    This means 3x+\pi \neq k \pi \Rightarrow 3x \neq k \pi (since k is any integer)
    So it's not defined for x=\frac{k\pi}{3}, for any integer k.

    The domain is then D=\mathbb{R}- \left\{\tfrac{k \pi}{3} ~:~ k \in \mathbb{Z}\right\}

    Does it look clear ?




    For the range... the cosecant's range is \mathbb{R}-(-1,1) (see here : http://mathworld.wolfram.com/Cosecant.html )
    So for any z in the domain of the cosecant, csc(z) \geq1 or csc(z) \leq-1. Thus 3csc(z) \geq3 or 3csc(z) \leq-3. --> 3csc(z)-2 \geq1 or 3csc(z) \leq-5.

    The range of 3 \csc(3x+\pi)-2 is hence \mathbb{R}-(-5,1)




    For the graph, I have no particular method... I'm sorry, because we never learnt csc so I can't help you more than your book on this :s
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  5. #5
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by Moo View Post
    ša va, merci

    \csc(z)=\frac{1}{\sin(z)}
    So it's not defined if \sin(z)=0.
    \sin(z)=0 \Leftrightarrow z=k \pi ~,~ k \in \mathbb{Z} (the set of all integers)

    So for \csc(3x+\pi) to be defined, you need 3x+\pi to be different from k \pi
    This means 3x+\pi \neq k \pi \Rightarrow 3x \neq k \pi (since k is any integer)
    So it's not defined for x=\frac{k\pi}{3}, for any integer k.

    The domain is then D=\mathbb{R}- \left\{\tfrac{k \pi}{3} ~:~ k \in \mathbb{Z}\right\}

    Does it look clear ?




    For the range... the cosecant's range is \mathbb{R}-(-1,1) (see here : Cosecant -- from Wolfram MathWorld )
    So for any z in the domain of the cosecant, csc(z) \geq1 or csc(z) \leq-1. Thus 3csc(z) \geq3 or 3csc(z) \leq-3. --> 3csc(z)-2 \geq1 or 3csc(z) \leq-5.

    The range of 3 \csc(3x+\pi)-2 is hence \mathbb{R}-(-5,1)




    For the graph, I have no particular method... I'm sorry, because we never learnt csc so I can't help you much than your book on this :s
    wow you are good...i think you learn it in a more complex [i think] method...but for range i think i got it ^_^

    for domain, i see that csc is undefined every xpi , and we have 3x + pi... but i would think to put them equal to each other, but it doesnt work out...would i just have to ignore the pi, or is that just wrong?

    muchas gracias por haberme ayudado =]
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  6. #6
    Moo
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    wow you are good...i think you learn it in a more complex [i think] method...but for range i think i got it ^_^
    Well, it was not a particular complex method, I just thought it would be clearer to see it this way...
    No sÚ como tu has aprendido :P

    for domain, i see that csc is undefined every xpi , and we have 3x + pi... but i would think to put them equal to each other, but it doesnt work out...would i just have to ignore the pi, or is that just wrong?
    Yes, you can ignore pi.
    Let 3x+pi=k pi (not xpi, it can be confusing because you already have one)
    It's equivalent to 3x=k pi, because k is any integer, so it doesn't matter whether you consider k or k-1.

    muchas gracias por haberme ayudado =]
    de nada ?
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  7. #7
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by Moo View Post
    Well, it was not a particular complex method, I just thought it would be clearer to see it this way...
    No sÚ como tu has aprendido :P


    Yes, you can ignore pi.
    Let 3x+pi=k pi (not xpi, it can be confusing because you already have one)
    It's equivalent to 3x=k pi, because k is any integer, so it doesn't matter whether you consider k or k-1.


    de nada ?
    Pues La manera que me estas ensenando es mas diferente de lo que nos ensenan en clase

    ok, now i get it...i mean i get blurry up to a point, but once i get past it, everything comes back to me quick...

    Muchisimas Gracias Senorita
    Haber Cuando Nos Hablamos eh?
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