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Math Help - Finding the minimum value

  1. #1
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    Finding the minimum value

    Hey there I need some help with this problem

    Find the minimum value of
    absolute value of sinx + cosx + tanx + cotx + secx + cscx

    I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Finding the minimum value

    Quote Originally Posted by gfbrd View Post
    Hey there I need some help with this problem

    Find the minimum value of
    absolute value of sinx + cosx + tanx + cotx + secx + cscx

    I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks.
    The derivative of:

    f(x) = \sin(x)+ \cos(x) + \tan(x) + \cot(x) + \sec(x) + \csc(x)

    is:

    \frac{df}{dx} = \cos(x)-\sin(x)+\tan^{2}(x)-\cot^{^2}(x)+\sec(x)  \tan(x)-\csc(x) \cot(x)


    The \frac{df}{dx} above can be written as(for y > 0):

    \begin{align*}=& \cos(x) - \sin(x) + \frac{\sin^{2}(x)}{\cos^{2}(x)} -\frac{\cos^{2}(x)}{\sin^{2}(x)} + \left(\frac{\sin(x)}{\cos(x)}\right)\left(\frac{1}  {\cos(x)}\right) - \left( \frac{\cos(x)}{\sin(x)}\right)\left(\frac{1}{\sin(  x)}\right) \\ \\ =& \cos(x) - \sin(x) + \frac{\sin^{2}(x)}{\cos^{2}(x)} -\frac{\cos^{2}(x)}{\sin^{2}(x)} + \frac{\sin(x)}{\cos^{2}(x)} - \frac{\cos(x)}{\sin^{2}(x)} \\ \\ =& \cos(x) - \sin(x) + \frac{\sin^{4}(x) - \cos^{4}(x)}{\sin^{2}(x) \cos^{2}(x)} + \frac{\sin^{3}(x) - \cos^{3}(x)}{\sin^{2}(x) \cos^{2}(x)} \\  \\ =& \frac{\sin^{2}(x) \cos^{2}(x)[\cos(x) - \sin(x) + \sin^{2}(x) - \cos^{2}(x) + \sin(x) - \cos(x)]}{\sin^{2}(x) \cos^{2}(x)}\\  \\ =& \cos(x) - \sin(x) + \sin^{2}(x) - \cos^{2}(x) + \sin(x) - \cos(x)\\ \\ =& -\cos(2x)\end{align*}


    Now to find absolute min set \frac{df}{dx} = 0 and solve for x:

    \begin{align*}-\cos(2x) =& 0 \\  2x =& \cdots, -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\ x =& \cdots, -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}, \cdots\end{align*}

    same for y < 0, \frac{df}{dx} will be:

    \frac{df}{dx} = \cos(2x) = 0

    So plugin the value of x one by one and test which one's the smallest. You can see that there's a pattern.


    Absolute min value is: f(\frac{3\pi}{4}) = 2

    Hope it helps.
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  3. #3
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    Re: Finding the minimum value

    Thanks a lot for your help
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  4. #4
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    Re: Finding the minimum value

    f(2.65345) = -1.82843 (5dp).
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  5. #5
    Senior Member x3bnm's Avatar
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    Re: Finding the minimum value

    Quote Originally Posted by BobP View Post
    f(2.65345) = -1.82843 (5dp).
    BobP you're right. Just curious can you kindly tell me how you got that mathematically?
    Last edited by x3bnm; November 30th 2012 at 04:32 AM.
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  6. #6
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    Re: Finding the minimum value

    Newton-Raphson on the derivative, (I didn't fancy trying to solve derivative = 0 algebraically). First approximation came from looking at the graph.
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  7. #7
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    Re: Finding the minimum value

    w8 the d(tanx)/dx is tan^2(x)? lol and the derivative of cot(x)=-cot^2(x)... i cant seem to follow that.....
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  8. #8
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    Re: Finding the minimum value

    Well I know the answer now is 2sqrt(2)-1
    Not sure how to get that though =/

    derivative of tan is sec^2 and
    derivative of cot is csc^2
    Last edited by gfbrd; November 30th 2012 at 05:25 PM.
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  9. #9
    Senior Member x3bnm's Avatar
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    Re: Finding the minimum value

    Sorry I messed up in my early post. Please disregard my last answer. That's absolutely garbage. Sorry about that.
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