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Math Help - sinx + tanx > 2x

  1. #1
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    sinx + tanx > 2x - help!!

    How do I show that sinx + tanx > 2x if 0 < x < pi/2 ?
    Last edited by weasley74; March 6th 2008 at 01:11 PM.
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  2. #2
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    Quote Originally Posted by weasley74 View Post
    How do I show that sinx + tanx > 2x if 0 < x < pi/2 ?
    You want to show 2x is the minimum boundary. Find the derivative of sin(x)+tan(x) and set it equal to zero. Solve for x and find the minimum.
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    so I get cosx + 1/(cosx)^2 = 0 , and then I get (cosx)^3 + 1 = 0, putting cosx = t, I get t^3 + 1 = 0

    and because what must be a minor stroke, i have no idea how to solve that simple equation.
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  4. #4
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    Quote Originally Posted by weasley74 View Post
    so I get cosx + 1/(cosx)^2 = 0 , and then I get (cosx)^3 + 1 = 0, putting cosx = t, I get t^3 + 1 = 0

    and because what must be a minor stroke, i have no idea how to solve that simple equation.
    Haha, t=-1
    x=\pi

    That is not in the range, so we are at a crossroads.
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  5. #5
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    so we didn't really find anything out by doing that?
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  6. #6
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    Both functions are non negative on the specified interval. The derivative of 2x is 2 while the derivative of sin(x) + tan(x) = cos(x)+1/cos(x)^2 which is clearly greater than or equal to 2. Since sin(x)+tan(x) = 2x at x=0 and sin(x)+tan(x) is increasing faster than 2x for every value of x (0,pi/2], sin(x)+tan(x) is greater than or equal to 2x.
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