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Math Help - Simple trigonometric inequality, yet no simple solution.

  1. #1
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    Simple trigonometric inequality, yet no simple solution.

    Hey guys!

    I've recently encountered this problem and I would like to know if any of you could present a more elegant or simple solution to it.

    $$cosx+tgx>sinx+1$$

    My solution:

    When $cosx\neq0$

    $cosx+$sinx\over cosx > $1+sinx$

    cos^2x+sinx-cosx-sinxcosx\over cosx >0

    {(cosx-1)(cosx-sinx)\over cosx}>0

    {\forall_x } {cosx-1\leq 0} So for cosx\neq 1 we have:

    {(cosx-sinx)\over cosx}<0

    {1-tgx}<0

    1<tgx We sketch the graph, find solutions and then see what happens if cosx=1
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  2. #2
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    Re: Simple trigonometric inequality, yet no simple solution.

    Hello, Doubled144314!

    I will limit my answers to Quadrant 1.
    You can generalize the results if you like.


     \cos x+\tan x \:>\:\sin x+1
    We note that x \ne \tfrac{\pi}{2}

    We have: . \cos x + \frac{\sin x}{\cos x} \:>\:\sin x + 1

    Multiply by \cos x\!:\;\;\cos^2\!x + \sin x \:>\:\sin x\cos x + \cos x

    . . . . \sin x - \sin x\cos x - \cos x + \cos^2x \:>\:0

    . . . . \sin x(1-\cos x) - \cos x(1-\cos x) \:>\:0

    . . . . . . . . . . (1-\cos x)(\sin x - \cos x) \:>\:0


    There are two cases: . \begin{bmatrix}\text{(1) both factors are negative} \\ \text{(2) both factors are positive}\end{bmatrix}


    Case (1): . 1-\cos x \:<\:0 \quad\Rightarrow\quad \cos x \:>\:1\;\text{ impossible}


    Case (2): . 1 - \cos \:>\:0 \quad\Rightarrow\quad \cos x\:<\:1 \quad\Rightarrow\quad x \in \left(0,\,\tfrac{\pi}{2}\right]

    . . . . . . . . \sin x-\cos x \:>\:0 \quad\Rightarrow\quad \sin x \:>\:\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:>\:1
    . . . . . . . . \tan x \:>\:1 \quad\Rightarrow\quad x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)


    Therefore: . x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)
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