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

• 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.