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Math Help - How does this algebraic work?

  1. #1
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    How does this algebraic work?

    In a book I am reading I have:

    \frac{xcosx + 3sinx}{cos^2 x}

    which gets transformed to:

    \frac{x + 3sinx}{cos x}

    How is that possible? Can someone please explain how that works?

    Angus
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  2. #2
    Junior Member BobBali's Avatar
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    Re: How does this algebraic work?

    Let \theta = x

    \frac{\theta \cdot cos\theta + 3sin\theta}{cos\theta \cdot cos\theta}

    Top and bottom Cos theta's cancel out leaving you with the answer
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  3. #3
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    Re: How does this algebraic work?

    Quote Originally Posted by BobBali View Post
    Let \theta = x

    \frac{\theta \cdot cos\theta + 3sin\theta}{cos\theta \cdot cos\theta}

    Top and bottom Cos theta's cancel out <--- No! you can't cancel summands!
    leaving you with the answer
    If you want to cancel a factor you have to factor denominator and numerator:

    \frac{\theta \cdot \cos(\theta) + 3 \sin(\theta)}{\cos(\theta) \cdot \cos(\theta)} = \frac{\cos(\theta) \cdot \left(\theta  + 3 \cdot \frac{\sin(\theta)}{\cos(\theta)}\right)}{ \cos(\theta) \cdot \cos(\theta)}
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  4. #4
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    Re: How does this algebraic work?

    Hello, angypangy!

    \text{In a book I am reading I have: }\:\frac{x\cos x + 3\sin x}{\cos^2\!x}

    . . \text{which gets transformed to: }\:\frac{x + 3\sin x}{\cos x}

    \text{How is that possible?}

    It is not possible.

    The choices are obvious:
    . . you misread the statement,
    . . the book has a typo,
    . . or the author is a fulldecker.

    Someone cancelled illegally: .  \frac{x\rlap{\:////}\cos x +3\sin x}{\cos^{\rlap{/}2}x} \;=\; \frac{x+3\sin x}{\cos x}\;\;??

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  5. #5
    Member sbhatnagar's Avatar
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    Re: How does this algebraic work?

    Yeah, maybe there's a typo.

    I am sure the author meant to say:

    \frac{x\cdot \cos{x}+3\sin{x}}{\cos^2{x}}= \frac{x +3\tan{x}}{\cos{x}}
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  6. #6
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    Re: How does this algebraic work?

    Quote Originally Posted by sbhatnagar View Post
    Yeah, maybe there's a typo.

    I am sure the author meant to say:

    \frac{x\cdot \cos{x}+3\sin{x}}{\cos^2{x}}= \frac{x +3\tan{x}}{\cos{x}}


    Sorry all, sbhatnager is correct, yes I meant to say
    \frac{x +3\tan{x}}{\cos{x}}

    Angus
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  7. #7
    MHF Contributor Amer's Avatar
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    Re: How does this algebraic work?

    Quote Originally Posted by BobBali View Post
    Let \theta = x

    \frac{\theta \cdot cos\theta + 3sin\theta}{cos\theta \cdot cos\theta}

    Top and bottom Cos theta's cancel out leaving you with the answer
    what is the point in your assumption  x = \theta ??
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  8. #8
    Member sbhatnagar's Avatar
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    Re: How does this algebraic work?

    Quote Originally Posted by angypangy View Post
    Sorry all, sbhatnager is correct, yes I meant to say
    \frac{x +3\tan{x}}{\cos{x}}

    Angus
    \frac{x\cdot \cos{x}+\sin{x}}{\cos^2{x}}

    Multiply the fraction by {\sec{x} \over \sec{x}}:
    \frac{x\cdot \cos{x}+\sin{x}}{\cos^2{x}}= \frac{x\cdot \cos{x} \cdot \sec{x} +\sec{x}\cdot\sin{x}}{\sec{x} \cdot\cos^2{x}}

    Put \sec{x}={1 \over \cos{x}}
    = \frac{x \cdot \left( {\cos{x} \over \cos{x}}\right)+{\sin{x} \over \cos{x}}}{\left( {\cos^2{x} \over \cos{x} }\right)} = \frac{x +{\sin{x} \over \cos{x}}}{\left( {\cos{x}}\right)}

    Put {\sin{x} \over \cos{x}}=\tan{x}
    =\frac{x +3\tan{x}}{\cos{x}}
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