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Math Help - What is wrong with this one?

  1. #1
    Junior Member
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    What is wrong with this one?

    \cos(4v)=-\frac{1}{\sqrt{2}} \Leftrightarrow
    -\cos(4v)=\frac{1}{\sqrt{2}} \Leftrightarrow
    cos(4v+\pi)=\frac{1}{\sqrt{2}} \Leftrightarrow
    4v+\pi=+-\frac{\pi}{4}+n2\pi
    I:
    4v+\pi=\frac{\pi}{4}+n2\pi \Leftrightarrow
    v=-\frac{3\pi}{16}+\frac{n\pi}{2}
    II:
    4v+\pi=-\frac{\pi}{4}+n2\pi \Leftrightarrow
    v=-\frac{5\pi}{16}+\frac{n\pi}{2}

    Where I and II are my two solutions.
    The answer given in my book are:
    +-\frac{3\pi}{16}+\frac{n\pi}{2}

    In this post I use a plus and a minus following eachother (like this: +-) to indicate that the following term can be either negative or positive.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by a4swe
    \cos(4v)=-\frac{1}{\sqrt{2}} \Leftrightarrow
    -\cos(4v)=\frac{1}{\sqrt{2}} \Leftrightarrow
    cos(4v+\pi)=\frac{1}{\sqrt{2}} \Leftrightarrow
    4v+\pi=+-\frac{\pi}{4}+n2\pi
    I:
    4v+\pi=\frac{\pi}{4}+n2\pi \Leftrightarrow
    v=-\frac{3\pi}{16}+\frac{n\pi}{2}
    II:
    4v+\pi=-\frac{\pi}{4}+n2\pi \Leftrightarrow
    v=-\frac{5\pi}{16}+\frac{n\pi}{2}

    Where I and II are my two solutions.
    The answer given in my book are:
    +-\frac{3\pi}{16}+\frac{n\pi}{2}

    In this post I use a plus and a minus following eachother (like this: +-) to indicate that the following term can be either negative or positive.
    There is nothing wrong with your solution, its just another way of expressing
    the solution in the book.

    Note:

    <br />
-\frac{5 \pi}{16}=\frac{3 \pi}{16} -\frac{\pi}{2}<br />

    RonL
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  3. #3
    Junior Member
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    Yes of course.
    I thought I had missed something fundamental, that was not the case.
    Thank you.
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  4. #4
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    Hello, a4swe!

    Just a question: why are you moving that minus-sign around?

    We have: . \cos(4v)\:=\:-\frac{1}{\sqrt{2}}\quad\Rightarrow\quad 4v \:= \:\pm\frac{3\pi}{4} + 2\pi n \quad\Rightarrow\quad v \:=\:\pm\frac{3\pi}{16} + \frac{\pi}{2}n

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  5. #5
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    To show (for my self maninly) that what I do is indeed right.
    I included it here just because I thought it was possible that it was somewhere there I did something wrong.
    If someone would say, hey there is your fault!
    Then I colud more easily trace it if I did it like I did it rather than how you did it, don't you think?

    Conclusion: I do it because I am a little rusty in trigonometry.
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  6. #6
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Soroban


    We have: . \cos(4v)\:=\:-\frac{1}{\sqrt{2}}\quad\Rightarrow\quad 4v \:= \:\pm\frac{3\pi}{4} + 2\pi n \quad\Rightarrow\quad v \:=\:\pm\frac{3\pi}{16} + \frac{\pi}{2}n

    How do you get rid of the cos sign? is there something you multiply by?
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  7. #7
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    Quote Originally Posted by Quick
    How do you get rid of the cos sign? is there something you multiply by?
    Yes, the inverse cosine function.
    Like here,
    x^2=2 how do you remove the square.
    You use the inverse function \sqrt{x} which removes it.
    ---
    However, over here that value is one of the trigonometric values that you need to have memorized.
    Since,
    \cos \pi/4=\sqrt{2}/2=1/\sqrt{2}
    So when it is negative, it is either in the 2nd or 3rd quadrant so,
    \pi-\pi/4
    \pi+\pi/4
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