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Math Help - 2sinx-cotxcosx=1

  1. #1
    Junior Member jonnygill's Avatar
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    2sinx-cotxcosx=1

    the problem is...

    2sin\Theta-cot\Theta cos\Theta=1

    after switching terms around and doing some operations i got...

    4sin^2\Theta=1+\dfrac{2}{sin^3\Theta}+\dfrac{1}{si  n^6\Theta}

    This feels like the wrong direction since i'm using terms with sin^6 in the denominator.

    When i first looked at it, it seemed like getting all the terms to be sin (or cos too i guess) would be possible. Well i got all the terms to be in sin, but these terms have large exponents in the denominator, etc.

    Is there another, easier approach?

    thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jonnygill View Post
    the problem is...

    2sin\Theta-cot\Theta cos\Theta=1
    Made an oopsie. Correct answer is already given so I'm not going to fix it.

    -Dan
    Last edited by topsquark; February 24th 2011 at 01:34 PM. Reason: Made an oopsie
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  3. #3
    Junior Member jonnygill's Avatar
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    Quote Originally Posted by topsquark View Post
    \displaystyle 2 sin( \theta ) - cot( \theta) cos( \theta ) = 0

    \displaystyle 2 sin( \theta ) - \frac{cos^2( \theta)}{sin(\theta )} = 0

    \displaystyle 2 sin^2( \theta ) - cos^2( \theta) = 0

    Now use \displaystyle sin^2( \theta ) = 1 - cos^2( \theta )

    -Dan

    could you show how you went from line 2 to line 3?

    i tried getting a common denominator but i ended up with -\dfrac{1}{sin^3\Theta}+2sin\Theta

    also, where did the 1 go in the original equation? why did you just remove it?


    edit: i see how you went from line 2 to line 3. but i do not understand why you set the equation equal to zero and not one. was this an oversight?
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  4. #4
    Junior Member jonnygill's Avatar
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    i kept the one and got 3sin^2\Theta-sin\Theta-1=0

    I can now use the quad. formula to solve for sin(Theta). I suspect you removed the 1 on purpose. perhaps it makes it easier to solve. but how would you re-incorporate that 1 back into the equation when you are finished?


    edit: i solved the equation using the quad. formula and got two answers that work with the original equation. Thanks!
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  5. #5
    MHF Contributor
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    Quote Originally Posted by jonnygill View Post
    the problem is...

    2sin\Theta-cot\Theta cos\Theta=1

    after switching terms around and doing some operations i got...

    4sin^2\Theta=1+\dfrac{2}{sin^3\Theta}+\dfrac{1}{si  n^6\Theta}

    This feels like the wrong direction since i'm using terms with sin^6 in the denominator.

    When i first looked at it, it seemed like getting all the terms to be sin (or cos too i guess) would be possible. Well i got all the terms to be in sin, but these terms have large exponents in the denominator, etc.

    Is there another, easier approach?

    thanks!
    Simply use

    \displaystyle\frac{cos\Theta}{sin\Theta}=cot\Theta

    which gives

    \displaystyle\ 2sin\Theta-\frac{cos\Theta}{sin\Theta}\;cos\Theta=1\Rightarro  w\left[\frac{sin\Theta}{sin\Theta}\right]2sin\Theta-\frac{cos^2\Theta}{sin\Theta}=1

    which gives a common denominator


    \displaystyle\frac{2sin^2\Theta-cos^2\Theta}{sin\Theta}=1

    2sin^2\Theta-\left(1-sin^2\Theta\right)=sin\Theta
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