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Math Help - i.d. proofs

  1. #1
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    i.d. proofs

    Hi:
    I am asked to prove the following identities:
    a) 2cot2θ≡cotθ-tanθ
    b) tan3θ≡3tanθ-tan³θ/1-3tan²θ

    I think I should tackle the most complicated side as it is easier to aim for the easier looking side. (hope that makes sense )
    I should warn you that I'm not good at this so please excuse my fumbling
    a) 2cot2θ≡cotθ-tanθ
    I will attempt the RHS, cotθ-tanθ
    cotθ-sinθ/cosθ
    so I'm stuck! Told you
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  2. #2
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    Quote Originally Posted by stealthmaths View Post
    Hi:
    I am asked to prove the following identities:
    a) 2cot2θ≡cotθ-tanθ
    b) tan3θ≡3tanθ-tan³θ/1-3tan²θ

    I think I should tackle the most complicated side as it is easier to aim for the easier looking side. (hope that makes sense )
    I should warn you that I'm not good at this so please excuse my fumbling
    a) 2cot2θ≡cotθ-tanθ
    I will attempt the RHS, cotθ-tanθ
    cotθ-sinθ/cosθ
    so I'm stuck! Told you
    I'd work from the left to avoid the double angle. It seems you are unaware that \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}

    \tan (2 \theta) = \frac{2\tan \theta}{1- \tan ^2 \theta}

    2\cot (2 \theta) = \frac{2}{\tan(2 \theta)} = \frac{2}{\frac{2\tan \theta}{1- \tan ^2 \theta}} = \frac{2(1-\tan^2 \theta)}{2 \tan \theta} = \frac{1-\tan ^2 \theta}{\tan \theta}

    Splitting the fraction:

    \frac{1}{\tan \theta} - \frac{\tan ^2 \theta}{\tan \theta} = \cot \theta - \tan \theta = \text{ RHS }
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  3. #3
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    re:reply

    Hi:
    thank you:
    I was aware of cotθ=1/tanθ=cosθ/sinθ
    But I find this a mine field. And never know where to start nor which replacements to use. It's really confusing to me.
    I would like to return to this a little later and ask you some questions about your thinking, if I may?

    But first:
    b) tan3θ≡(3tanθ-tan³θ)/1-3tan²θ
    so the LHS has only one thing to play with. So I should tackle this side?

    tan3θ=sin3θ/cos3θ

    On the right track?
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  4. #4
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    Quote Originally Posted by stealthmaths View Post
    Hi:
    thank you:
    I was aware of cotθ=1/tanθ=cosθ/sinθ
    But I find this a mine field. And never know where to start nor which replacements to use. It's really confusing to me.
    I would like to return to this a little later and ask you some questions about your thinking, if I may?

    But first:
    b) tan3θ≡(3tanθ-tan³θ)/1-3tan²θ
    so the LHS has only one thing to play with. So I should tackle this side?

    tan3θ=sin3θ/cos3θ

    On the right track?
    In my opinion, for b), tackle the left hand side by writing tan3\theta=(tan2\theta+\theta) and use the addition formulae.
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Punch View Post
    In my opinion, for b), tackle the left hand side by writing tan3\theta=(tan2\theta+\theta) and use the addition formulae.
    I agree.

    Also thank goodness for LaTeX. It's not that hard to write on paper though

    \tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B}

    \therefore \: \: \tan (2 \theta) = \frac{2 \tan \theta}{1-\tan ^2 \theta}

    Write \tan (3 \theta) as \tan (2 \theta + \theta)

    Note that A = 2 \theta and B = \theta

    Hence \tan (2 \theta + \theta) = \frac{\tan (2 \theta) + \tan \theta}{1- \tan (2 \theta) \tan \theta}

    Subbing in our double angle identity for \tan wherever \tan (2 \theta) occurs:

    \frac{\frac{2 \tan \theta}{1-\tan ^2 \theta}+\tan \theta}{1 - \frac{2 \tan \theta}{1-\tan ^2 \theta} \tan \theta}<br />



    Looking at the numerator:


    \frac{2 \tan \theta}{1-\tan ^2 \theta}+\tan \theta

    Get the same denominator by multipling \tan \theta by \frac{1-\tan^2 \theta}{1-\tan^2 \theta}

    \frac{2 \tan \theta + \tan \theta (1- \tan ^2 \theta)}{1-\tan^2 \theta}


    Distribute the \tan \theta and simplify:

    <br />
\frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta}


    Note the similarity of this to the right hand side's numerator. Therefore, leave it like this and work on the denominator




    Looking at the denominator (of the overall fraction)


    1 - \frac{2 \tan \theta}{1-\tan^2 \theta} \tan \theta


    Rewrite 1 as \frac{1-\tan^2 \theta}{1-\tan^2 \theta}

    The second term can be simplified to \frac{2 \tan ^2 \theta}{1-\tan^2 \theta}


    Since the two terms now have the same denominator we can combine them:


    \frac{(1- \tan^2 \theta)- 2 \tan^2 \theta}{1- \tan^2 \theta}

    The numerator of this fraction can be simplified to 1-3 \tan^2 \theta

    \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta}


    As this is also similar to the RHS leave it in this form.



    Bringing both parts together


    Numerator: \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta}

    Denominator:  \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta}


    Which makes our fraction equal to \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \div \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta}


    Dividing by a fraction is the same as 'flipping' it (taking the reciprocal) and multiplying.

    <br />
\frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \div \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta} = \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \times \frac{1- \tan^2 \theta}{1-3 \tan^2 \theta}


    1- \tan^2 \theta cancels to leave us with:

    \frac{3 \tan \theta - \tan^3 \theta}{1- 3 \tan^2 \theta} = \text{ RHS}
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  6. #6
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    Well. That was a rather bigger than expected process.
    Thanks for that

    Your right - I should learn Latex, and will make an effort to do so after this weekend, which is this coming Monday. Promise Promise.
    For anybody reading this who is interested, I just did a search on Latex video tutorials and came up with a couple of sites instantly. It seems there is a lot of help available online, as always.

    TechScreencast - LaTeX Intro :: tutorial, videos, articles, screencast to learn technology"

    latex video tutorial rapidshare, megaupload, torrent download

    Though I haven't looked closely at them, yet, I'm pretty certain they will be helpful and are reputable sites.
    Is the last link ok?
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