Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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).

share|cite|improve this answer

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

share|cite|improve this answer

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 ${ 22 \over 7} < \sqrt{12}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.