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Thread: proving sin^2(x)<=|sin(x)|

  1. #1
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    proving sin^2(x)<=|sin(x)|

    For all x, prove that $\displaystyle sin^{2}(x)\leq|sin(x)|$ given that x is in the set of real numbers.

    I think i need to break this into a couple different cases, specifically: (0,$\displaystyle \pi$),($\displaystyle \pi$,$\displaystyle 2\pi$), but am not sure. Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by snaes View Post
    For all x, prove that $\displaystyle sin^{2}(x)\leq|sin(x)|$ given that x is in the set of real numbers.
    Don’t make it difficult. Recall that $\displaystyle 0\leqslant a \leqslant 1\, \Rightarrow \,a^2 \leqslant a $.

    We know that $\displaystyle 0\le |\sin(x)|\le1$ so $\displaystyle \sin^2(x)=|\sin(x)|^2\le |\sin(x)|$.
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    Thanks, I was definitely over-thinking that.
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