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This is a two part question, and I need help with the 2nd part.

For this 1st question, I found the remainder term and showed that for $0 < c < \pi$, the remainder term is $> 0$, therefore, the inequality holds true. Is this sufficient proof?

The 2nd part of the question asks:

Show that the inequality in part (1) holds for all real positive x and deduce that:

$x-\frac{x^3}{6} < \sin x < x$

for all real positive $x$.

Can someone help me with this second part?

Thanks in advance!

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    Note that $x-x^3/3$ is decrasing. $x$ is tangent to the sine at x=0, and the sine is concave in $(0,\pi)$.2012-06-04

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Examine the derivatives. You can do something like this:

If we have that $f'(x)>0$ for every $x>a$, then $\int_a^x\!f'(t)\,dt>0$ which implies $f(x)>f(a)$ for every $x>a$. A similar result holds for $f'(x)<0$.

For $\sin(x) you clearly only need to worry about $x\in(0,1]$, and applying this to $x-\sin x$ will take care of that nicely. For the other inequality, you've already taken care of the hard part, now just apply this approach with $a=3$, for instance.

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    @mathstudent No problem. Also your approach to the first part is sufficient proof.2012-06-04