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

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September 22nd 2010, 12:00 PM
snaes

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

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

I think i need to break this into a couple different cases, specifically: (0, $\pi$ ),( $\pi$ , $2\pi$ ), but am not sure. Any help would be appreciated.
September 22nd 2010, 12:13 PM
Plato

Quote:

Originally Posted by snaes

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

Don’t make it difficult. Recall that $0\leqslant a \leqslant 1\, \Rightarrow \,a^2 \leqslant a$ .

We know that $0\le |\sin(x)|\le1$ so $\sin^2(x)=|\sin(x)|^2\le |\sin(x)|$ .
September 22nd 2010, 12:14 PM
snaes

Thanks, I was definitely over-thinking that.

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