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Math Help - Trig Identity

  1. #1
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    Trig Identity

    Prove the Identity:

    cot\;\theta - tan\;\theta = 2\;cot\;2\;\theta

    My Work:

    \frac{cos\;\theta}{sin\;\theta} - \frac{sin\;\theta}{cos\;\theta} = 2\;cot\;2\;\theta

    \frac{cos^2\;\theta - sin^2\;\theta}{sin\;\theta\;cos\;\theta} = 2\;cot\;2\;\theta

    What do I do after I have all of this done?
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  2. #2
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    See my response to your other post:

    http://www.mathhelpforum.com/math-he...rig-proof.html

    and take it to heart. Also Plato's step by step proof in the same thread offers an example of the correct way to do this type of problem.
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  3. #3
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    I did look at his proof. And so far I've done what he did in the last proof. But then it gets tricky and I need some help.
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  4. #4
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    Quote Originally Posted by OzzMan View Post
    Prove the Identity:

    cot\;\theta - tan\;\theta = 2\;cot\;2\;\theta

    My Work:

    \frac{cos\;\theta}{sin\;\theta} - \frac{sin\;\theta}{cos\;\theta} = 2\;cot\;2\;\theta

    \frac{cos^2\;\theta - sin^2\;\theta}{sin\;\theta\;cos\;\theta} = 2\;cot\;2\;\theta

    What do I do after I have all of this done?
    Recall the double angle formulae:

    \cos (2 \theta) = \cos^2 \theta - \sin^2 \theta

    \sin (2 \theta) = 2 \sin \theta \cos \theta

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  5. #5
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    \frac{cos^2\;\theta - sin^2\;\theta}{sin\;\theta\;cos\;\theta} = 2\;\frac{cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta}

    \frac{cos^2\;\theta - sin^2\;\theta}{sin\;\theta\;cos\;\theta} = \frac{cos^2 \theta - \sin^2 \theta}{sin \theta \cos \theta}

    This right?
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  6. #6
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    Notice that the right hand side of the equation you have a double-angle and on the right hand side you have a single angle \theta. This should give you a hint that you will probably have to use a double-angle identity. Perhaps:

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

    Or as mr. fantastic pointed out one of the many other double angle identities (his, in fact, is easier to apply).

    Arriving at the conclusion that 1 = 1 by starting with -1 = 1 and manipulting both sides of the equation is wrong.

    The proof should go like this:

    \cot(\theta)-\tan(\theta) (write down left hand side of equation)

     = \frac{\cos(\theta)}{\sin(\theta)} - \frac{\sin(\theta)}{\cos(\theta)} (by the fact that \tan(x) = \sin(x)/\cos(x) \text{ and } \cot(x) = 1/\tan(x) )

     = \frac{\cos^2(\theta)-\sin^2(\theta)}{\sin(\theta)\cos(\theta)} (by finding a common denominator and simplifying)

     = \frac{\cos(2\theta)}{1/2\sin(2\theta)} (by the identities mr. fantastic gave)

     = 2\cot(2\theta) (end with the right hand side of the equation)

    This is fundamentally different from the approach you took.
    Last edited by iknowone; March 18th 2008 at 05:52 PM.
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  7. #7
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    We don't always start with a complicated side.
    \begin{array}{l}<br />
 2\cot (2\theta ) \\ <br />
 2\frac{{\cos (2\theta )}}{{\sin (2\theta )}} \\ <br />
 \frac{{\cos ^2 (\theta ) - \sin ^2 (\theta )}}{\begin{array}{l}<br />
 \sin (\theta )\cos (\theta ) \\ <br />
 \frac{{\cos (\theta )}}{{\sin (\theta )}} - \frac{{\sin (\theta )}}{{\cos (\theta )}} \\ <br />
 \end{array}} \\ <br />
 \end{array}<br />
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  8. #8
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    Well. Whose method should I go with. Yours or what I posted?
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  9. #9
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    Doesn't matter. They both work. You proved that one side is equivalent to the other. Doesn't matter which side you start with. iknowone started with the left, Plato started with the right. Same result in the end.
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  10. #10
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    Quote Originally Posted by iknowone View Post

     = \frac{\cos^2(\theta)-\sin^2(\theta)}{\sin(\theta)\cos(\theta)} (by finding a common denominator and simplifying)

     = \frac{\cos(2\theta)}{2\sin(2\theta)} (by the identities mr. fantastic gave)

     = 2\cot(2\theta) (end with the right hand side of the equation)
    I'm a little confused because I'm not sure where you got that 2 from in  \frac{\cos(2\theta)}{2\sin(2\theta)}

    I get that cos\;2\;\theta = cos^2\;\theta - sin^2\;\theta

    But sin\;\theta\;cos\;\theta = sin\;2\;\theta So I'm little confused where that 2 infront of sin\;2\;\theta comes from. If anyone could shed some light on this that be great.
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  11. #11
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    It was a mere typo. What he meant was:

    \frac{cos^{2}\theta - sin^{2}\theta}{sin\theta cos\theta}

    = \frac{cos(2\theta)}{\frac{1}{2}\left(2 sin\theta cos\theta\right)}

    = \frac{cos(2\theta)}{\frac{1}{2}sin(2\theta)}
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  12. #12
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    Quote Originally Posted by o_O View Post
    It was a mere typo. What he meant was:

    \frac{cos^{2}\theta - sin^{2}\theta}{sin\theta cos\theta}

    = \frac{cos(2\theta)}{\frac{1}{2}\left(2 sin\theta cos\theta\right)}

    = \frac{cos(2\theta)}{\frac{1}{2}sin(2\theta)}
    Where does the \frac{1}{2} come from?
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  13. #13
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    I really hate to say this but it is absolutely true.
    All of this could have been avoided with a good grounding in middle school algebra.
    Friends, That is a sad commentary!
    If one does not understand the basics, one should not expect to be a player.
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  14. #14
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    You know that \sin(2\theta) = 2\sin\theta \cos\theta.

    You have \sin\theta \cos\theta in your bottom expression which resembles the expression above except you're missing that factor of 2. So if you want to use that factor of 2, you must counter it by multiply by 1/2 (since 1/2 x 2 = 1).

    To see it algebraically, divide both sides by 2 of the first expression to get:

    \frac{1}{2}\sin(2 \theta) = \sin\theta\cos\theta
    \frac{1}{2} \left(2 \sin\theta \cos \theta\right) = \sin\theta\cos\theta
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  15. #15
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    To me it just seems like that number just appears there without question. That is what I'm saying. I don't think it goes back to algebra, because I am very good algebra. I'm still confused but I will agree with what you are saying. But really all I was saying was I do not get how you go from this step <br />
\frac{cos^{2}\theta - sin^{2}\theta}{sin\theta cos\theta} to this \frac{cos(2\theta)}{\frac{1}{2}\left(2 sin\theta cos\theta\right)} I know that you use identities. But it randomly seems like a 2 and 1/2 are put in there for no reason. Just seems to me as this way to go about proving the identity is more complicated then what I said, which in turn is confusing me which is right and which is wrong.
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