# Proving that $\sin x \gt \dfrac x2$

I was working on this question and I got a contradiction.

$\sin x \gt \dfrac x2$ for $0 \lt x \lt \dfrac {\pi}{2}$

$\arccos ( \sin x)) \gt \arccos (\dfrac x2)$

$\dfrac {\pi}{2} -x \gt \arccos (\dfrac x2)$

$\arcsin (\dfrac x2) +\arccos (\dfrac x2) -x \gt \arccos (\dfrac x2)$

$\arcsin (\dfrac x2) \gt x$

$\dfrac x2 >\sin x$

Why am I getting this contradiction if the original statement is true? Thanks.

P.S. I evaluated $\arccos ( \sin x))$ using Wolframalpha.

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Note that $\cos^{-1}\left(x\right)$ is decreasing for $0\leq x\leq1$. Therefore, $\sin\left(x\right) > \frac{x}{2}$ implies that $\cos^{-1}\left(\sin\left(x\right)\right)<\cos^{-1}\left(\frac{x}{2}\right)$ (the inequality symbol flips).

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The function $\arccos(x)$ is decreasing in the interval $0\le x\le 1$.

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Note that for $x\ge 0$, we have $\sin x \ge x - { x^3 \over 3!}$, or for $x \neq 0$, ${\sin x \over x} \ge 1 -{x^2 \over 3!}$. The latter quantity is strictly greater than ${1 \over 2}$ when $|x|<\sqrt{3}$, and since $\sqrt{3} > { \pi \over 2}$, we have the desired result.

As an aside, it is straightforward to verify that $\pi < { 22 \over 7}$ (see good old Wikipedia), and that ${ 22 \over 7} < \sqrt{12}$.

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I don't really see how the result follows. As far as I can see, you proved that the RHS is positive. But the LHS is also positive on the interval, so how does that say anything about which side is larger? – Ovi Feb 13 '14 at 5:28
@Ovi: Thanks for catching that. I meant greater than a half, not positive. – copper.hat Feb 13 '14 at 5:31