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Math Help - Factor Formula

  1. #1
    Member SengNee's Avatar
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    Question Factor Formula

    When my textbook proving \frac{d}{dx}(sin\theta)=cos\theta, it mentions about "Factor Formula", but it never mention it detailly, so I don't understand it.
    Can anyone tell me what is "Factor Formula"?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post
    When my textbook proving \frac{d}{dx}(sin\theta)=cos\theta, it mentions about "Factor Formula", but it never mention it detailly, so I don't understand it.
    Can anyone tell me what is "Factor Formula"?
    No idea. Can you post the derivation? Perhaps it is different than the one I know.

    -Dan
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  3. #3
    Member SengNee's Avatar
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    Quote Originally Posted by topsquark View Post
    No idea. Can you post the derivation? Perhaps it is different than the one I know.

    -Dan

    To find \lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}


    Area of \bigtriangleup OAB < Area of sector OAB < Area of \bigtriangleup OAT
    0.5(1)^2(sin\theta)<0.5(1^2)(\theta)<0.5(1)(tan\th  eta)
    sin\theta<\theta<tan\theta
    1<\frac{\theta}{sin\theta}<\frac{1}{cos\theta}

    As \theta\rightarrow 0,

    cos\theta\rightarrow 1 and \frac{1}{cos\theta}\rightarrow 1.

    Therefore, \frac{\theta}{sin\theta}\rightarrow 1
    or
    \frac{sin\theta}{\theta}\rightarrow 1
    Hence
    \lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}=1


    Therefore, \frac{\theta}{sin\theta}\rightarrow 1
    or
    \frac{sin\theta}{\theta}\rightarrow 1
    Hence
    \lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}=1
    Actually, I don't understand this part, because
    1<\frac{\theta}{sin\theta}<\frac{1}{cos\theta}
    1<\frac{\theta}{sin\theta}<1
    What is this?
    (1,1)?
    Have range like this????
    Why \lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}=1 ?
    As \theta \rightarrow 0, the denominator is 0, it is undefined.




    Derivative of sinx

    Let f(x)=sinx where x is measured in radians.

    f(x)=sinx
    f(x+\delta x)=sin(x+\delta x)

    f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{f(x+\delta x)-f(x)}{\delta x}\right]}
    f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{sin(x+\delta x)-sin x}{\delta x}\right]}
    f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{2cos(x+\frac{1}{2}\delta x)sin\frac{1}{2}\delta x}{\delta x}\right]} (factor formula)
    f'(x)=\lim_{\delta x \rightarrow 0}{\left[cos(x+\frac{1}{2}\delta x)\cdot \frac{sin\frac{1}{2}\delta x}{\frac{1}{2}\delta x}\right]}
    f'(x)=cosx

    Since,
    \lim_{\delta x \rightarrow 0}{cos \left(x+\frac{1}{2}\delta x\right)}=cosx
    and
    \lim_{\delta x \rightarrow 0}{\left(\frac{sin\frac{1}{2}\delta x}{\frac{1}{2}\delta x}\right)}=1 as \left(\lim_{\delta x \rightarrow 0}{\frac{sin\theta}{\theta}}=1\right)

    So, what is factor formula???
    Last edited by SengNee; March 21st 2008 at 02:26 AM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post
    Actually, I don't understand this part, because
    1<\frac{\theta}{sin\theta}<\frac{1}{cos\theta}
    1<\frac{\theta}{sin\theta}<1
    What is this?
    This winds up being an application of something called the "squeeze theorem." You take two expressions, one that is less than the expression you want the limit for and one that is greater. Then if, as you apply your limit, the limit of the lesser and greater expressions are the same then the limit of your expression, the one in the middle, must be the same as these.

    So 1 is always less than \frac{\theta}{sin(\theta)} is always less than sec(\theta). As you take the limit as \theta \to 0 we find that
    1 < \frac{\theta}{sin(\theta)} < 1
    so
    \lim_{\theta \to 0}\frac{\theta}{sin(\theta)} = 1

    Quote Originally Posted by SengNee View Post
    f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{sin(x+\delta x)-sin x}{\delta x}\right]}
    f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{2cos(x+\frac{1}{2}\delta x)sin\frac{1}{2}\delta x}{\delta x}\right]} (factor formula)
    What your instructor is calling the "factor formula" I call by the name of a "sum to product" formula. By using the sum of angles formulas for sine and cosine
    sin(A \pm B) = sin(A)~cos(B) \pm sin(B)~cos(A)

    cos(A \pm B) = cos(A)~cos(B) \mp sin(A)~sin(B)

    we may construct the following formulas:
    sin(A) \pm sin(B) = 2~sin \left ( \frac{A \pm B}{2} \right )~cos \left ( \frac{A \mp B}{2} \right)

    cos(A) + cos(B) = 2~cos \left ( \frac{A + B}{2} \right )~cos \left ( \frac{A - B}{2} \right )

    cos(A) - cos(B) = -2~sin \left ( \frac{A + B}{2} \right )~sin \left ( \frac{A - B}{2} \right )

    For sin(x + \delta x) - sin(x) we may use the top expression with A = x + \delta x and B = x:
    sin(x + \delta x) - sin(x) = 2~sin \left ( \frac{(x + \delta x) - x)}{2} \right )~cos \left ( \frac{(x + \delta x) + x}{2} \right )

    = 2~sin \left ( \frac{\delta x}{2} \right )~cos \left ( x + \frac{\delta x}{2} \right )

    -Dan
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  5. #5
    Member SengNee's Avatar
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    Quote Originally Posted by topsquark View Post
    This winds up being an application of something called the "squeeze theorem." You take two expressions, one that is less than the expression you want the limit for and one that is greater. Then if, as you apply your limit, the limit of the lesser and greater expressions are the same then the limit of your expression, the one in the middle, must be the same as these.

    So 1 is always less than \frac{\theta}{sin(\theta)} is always less than sec(\theta). As you take the limit as \theta \to 0 we find that
    1 < \frac{\theta}{sin(\theta)} < 1
    so
    \lim_{\theta \to 0}\frac{\theta}{sin(\theta)} = 1


    What your instructor is calling the "factor formula" I call by the name of a "sum to product" formula. By using the sum of angles formulas for sine and cosine
    sin(A \pm B) = sin(A)~cos(B) \pm sin(B)~cos(A)

    cos(A \pm B) = cos(A)~cos(B) \mp sin(A)~sin(B)

    we may construct the following formulas:
    sin(A) \pm sin(B) = 2~sin \left ( \frac{A \pm B}{2} \right )~cos \left ( \frac{A \mp B}{2} \right)

    cos(A) + cos(B) = 2~cos \left ( \frac{A + B}{2} \right )~cos \left ( \frac{A - B}{2} \right )

    cos(A) - cos(B) = -2~sin \left ( \frac{A + B}{2} \right )~sin \left ( \frac{A - B}{2} \right )

    For sin(x + \delta x) - sin(x) we may use the top expression with A = x + \delta x and B = x:
    sin(x + \delta x) - sin(x) = 2~sin \left ( \frac{(x + \delta x) - x)}{2} \right )~cos \left ( \frac{(x + \delta x) + x}{2} \right )

    = 2~sin \left ( \frac{\delta x}{2} \right )~cos \left ( x + \frac{\delta x}{2} \right )

    -Dan

    How to prove them?

    \sin A\pm \sin B=2\sin\left(\frac{A\pm B}{2}\right)\cos\left(\frac{A\mp B}{2}\right)
    \cos A + \cos B=2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
    \cos A - \cos B=-2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A- B}{2}\right)

    \cos A\cos B=\frac{1}{2}\left[\cos\left(A+B\right)+\cos\left(A-B\right)\right]
    \sin A\sin B=-\frac{1}{2}\left[\cos\left(A+B\right)-\cos\left(A-B\right)\right]
    \cos A\sin B=\frac{1}{2}\left[\sin\left(A+B\right)-\sin\left(A-B\right)\right]
    \sin A\cos B=\frac{1}{2}\left[\sin\left(A+B\right)+\sin\left(A-B\right)\right]
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