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Math Help - Proving an identity.

  1. #1
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    Proving an identity.

    I keep coming to a dead end when working this one.

    csc x + cot x = \frac{1}{csc x - cot x}


    Thanks!
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  2. #2
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    Quote Originally Posted by JennyFlowers View Post
    I keep coming to a dead end when working this one.

    csc x + cot x = \frac{1}{csc x - cot x}


    Thanks!
    Work with the right hand side. Substitute csc x = 1/sin x and cot x = cos x/sin x. Simplify. Note that \displaystyle \frac{1}{1 - \cos (x)} = \frac{1 + \cos (x)}{\sin^2 (x)}. If you need more help, please show - clearly - all your work and say where you get stuck.
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  3. #3
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    Thank you. Here's what I have so far:

    \frac{1}{csc x - cot x} = \frac{1}{\frac{1}{sin x}-\frac{cos x}{sin x}} = \frac{1}{\frac{1 - cos x}{sin x}}=\frac{sin x}{1-cos x}

    I'm not sure what to do at this point.
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  4. #4
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    Divide everything by a cleverly disguised 1: \, \dfrac{\sin (x)}{\sin (x)}
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  5. #5
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    So that gives me:

    \frac{\frac{sin(x)}{1-cos(x)}}{\frac{sin(x)}{sin(x)}} = \frac{sin^2(x)}{(1-cos(x)) (sin(x))}

    I'm not quite sure if that's what you meant by "divide everything". This doesn't seem to help.

    Maybe...

    \frac{(1 - cos(x))(1 + cos(x))}{(1-cos(x)) (sin(x))} = \frac{1 + cos(x)}{sin(x)}

    Am I on the right track?
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  6. #6
    Super Member 11rdc11's Avatar
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    here is another way to approach it

    (\csc{x}+\cot{x})(\csc{x}-\cot{x}) = 1

    which is

    \csc^2{x}-\cot^2{x} =1

    Can you finish up from here?
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  7. #7
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    Quote Originally Posted by JennyFlowers View Post
    Thank you. Here's what I have so far:

    \frac{1}{csc x - cot x} = \frac{1}{\frac{1}{sin x}-\frac{cos x}{sin x}} = \frac{1}{\frac{1 - cos x}{sin x}}=\frac{sin x}{1-cos x}

    I'm not sure what to do at this point.
    What you do is go back and read what I noted in my previous post.
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  8. #8
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    Looks like I just about had it solved and didn't see it. As long as I'm right about Educated's tip:

    \frac{1 + cos(x)}{sin(x)} = csc x + cot x

    Did I do it right?
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  9. #9
    Super Member 11rdc11's Avatar
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    yep that is it
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    What you do is go back and read what I noted in my previous post.
    What part? Where you said to show my work and where I got stuck? I did show my work and where I got stuck...

    I also see where you said:

    Note that \displaystyle \frac{1}{1 - \cos (x)} = \frac{1 + \cos (x)}{\sin^2 (x)}
    I'm not sure how this is true or how I would apply it to the problem. Does it represent an identity?
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  11. #11
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    I see how Educated's tip solved the problem, I'm just not sure how I'd have figured that out on my own. It seems arbitrary to multiply everything by \frac{sin(x)}{sin(x)} until you actually do it.

    So if I were to receive this problem on a test, how would I know to do that? Identities are very frustrating for me. It often seems like trying random things until something works.

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  12. #12
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    Just a note: in LaTeX when you use \frac it gives you smaller fractions eg. \frac{\sin x}{\sin x}. This gets really hard to read if you have fractions on fractions. Use \dfrac to give bigger fractions eg. \dfrac{\sin x}{\sin x}

    Quote Originally Posted by JennyFlowers View Post
    \frac{\frac{sin(x)}{1-cos(x)}}{\frac{sin(x)}{sin(x)}} = \frac{sin^2(x)}{(1-cos(x)) (sin(x))}

    I'm not quite sure if that's what you meant by "divide everything".
    What I meant was \frac{\sin x}{1- \cos x} \div \frac{\sin x}{\sin x} which would give you: \dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}

    This gives you the answer straight away.


    Quote Originally Posted by JennyFlowers View Post
    I see how Educated's tip solved the problem, I'm just not sure how I'd have figured that out on my own. It seems arbitrary to multiply everything by \frac{sin(x)}{sin(x)} until you actually do it.

    So if I were to receive this problem on a test, how would I know to do that? Identities are very frustrating for me. It often seems like trying random things until something works.
    I noticed how close you were to proving the identity and realised that the numerator must be a 1. So if you divided the numerator by sin(x) it will give you 1.
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  13. #13
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    Quote Originally Posted by Educated View Post
    What I meant was \frac{\sin x}{1- \cos x} \div \frac{\sin x}{\sin x} which would give you: \dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}
    Strangely, when I work it this way, I don't get quite the same answer...

    \dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}=\dfrac{1}{\csc x - \cot x}

    That way doesn't seem to work, though I'm sure there's something I'm missing. I'd like to understand!
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  14. #14
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    Quote Originally Posted by JennyFlowers View Post
    What part? Where you said to show my work and where I got stuck? I did show my work and where I got stuck...

    I also see where you said:

    Note that

    I'm not sure how this is true or how I would apply it to the problem. Does it represent an identity?
    I don't think the missing step below should be too difficult to think of in this sort of topic:

    \displaystyle \frac{1}{1 - \cos (x)} \cdot \frac{1 + \cos (x)}{1 + \cos (x)} = ....

    Doing this is more suggestive (I would have thought) than the approach you ended up taking (which pretty much pulls a rabbit out of the hat). You do it as the next step from what you had way back in post #3.
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  15. #15
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    Quote Originally Posted by mr fantastic View Post
    I don't think the missing step below should be too difficult to think of in this sort of topic:

    \displaystyle \frac{1}{1 - \cos (x)} \cdot \frac{1 + \cos (x)}{1 + \cos (x)} = ....

    Doing this is more suggestive (I would have thought) than the approach you ended up taking (which pretty much pulls a rabbit out of the hat). You do it as the next step from what you had way back in post #3.
    Sorry, what I meant was that the last step in post #3 was:

    \frac{\sin (x)}{1- \cos (x)}

    I'm not sure how that is the same as where you're starting from:

    \frac{1}{1 - \cos (x)}
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