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Math Help - Inequality

  1. #1
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    Inequality

    Can somebody show me how to finish this one off? I know that the answer is [0,  \frac{\pi}{4}) and (\frac{5\pi}{4},2\pi], but I don't know how to lay this out. Here is what I have:

    \sin{x} > \cos{x}
    \sin{x} - \sin{(\frac{\pi}{2}-x}) > 0

    Not sure how to solve this algebraically.

    Thanks.
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  2. #2
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    Think about solving this \sin{x}=\cos{x}

    x=\frac{\pi}{4}+\pi k \ \ \ k\in\mathbb{Z}

    Now, if we restrict the solutions to the unit circle, we have x=\frac{\pi}{4} \ \ \frac{5\pi}{4}

    Now, test point to the left and right of those numbers.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Think about solving this \sin{x}=\cos{x}
    This is a good thought. It may be easier to see where dwsmith found he's solution from by the following

    \sin{x}=\cos{x}

    \frac{\sin{x}}{\cos{x}}=1

    \tan{x}=1\implies x=\frac{\pi}{4}+\pi k \ \ \ k\in\mathbb{Z}
    Last edited by pickslides; January 13th 2011 at 04:37 PM. Reason: bad latex..
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