I was doing a physics problem today and I needed to find when some movement is accelerating and when it is decelerating. So after some calculations, I got the equation I needed. I know that it is correct because when plotted, I can see from graph that I got the expected solution.

Unfortunately I don't know how to actually start solving the inequality. It looks like $$k \sin\left(\frac {\pi x} {2}\right)\cos\left(\frac {\pi x} {2}\right)>0$$ where $k$ is some huge sausage which is constant and positive.

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    $\begingroup$ Your identity is equivalent to $k\sin \pi x>0$ $\endgroup$ – Américo Tavares Sep 10 '10 at 20:30
  • $\begingroup$ @Arturo: I rolled back to a change of mine which was overridden by your rollback (the title only change). The title formula now matches the question. $\endgroup$ – Aryabhata Sep 10 '10 at 20:38
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    $\begingroup$ "sausage" ?!??! $\endgroup$ – Jason S Sep 11 '10 at 0:41
  • $\begingroup$ @M: Ehr... I'm not sure what that means. There seemed to be an attempt at editing the title that created a typo (the argument for the cosine was written as "p ix" instead of "pi x") which I tried to fix; my first attempt at editing someone else's post. $\endgroup$ – Arturo Magidin Sep 11 '10 at 0:50
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    $\begingroup$ @AndrejaKo: Ah, that kind of sausage. Yes, we tend to use it in the context of politics + shady manufacturers. The terminology is a bit different, though; not sure how to correct + keep the colloquialism. I would probably replace "sausage" with "expression". Less flavorful but clearer. :-) $\endgroup$ – Jason S Sep 14 '10 at 23:38

You can use the identity

$$\sin 2x = 2 \sin x \cos x$$

Thus you only need to determine when $$ \sin \pi x > 0$$ and this happens if and only if $$ x \in (2m, 2m+1), m \in \mathbb{Z}$$


Since $k$ is positive, it is immaterial. You want to know the values of $x$ for which $\sin(\frac{\pi x}{2})$ and $\cos(\frac{\pi x}{2})$ have the same sign.

Each function has period $4$, so we can just look at what happens on $[0,4]$. $\sin(\frac{\pi x}{2})$ is positive when $0\lt\frac{\pi x}{2}\lt\pi$ (that is, when $0\lt x\lt 2$), and negative when $2\lt x \lt 4$. $\cos(\frac{\pi x}{2})$ is positive when $0\leq \frac{\pi x}{2}\lt \frac{\pi}{2}$ and when $\frac{3\pi}{2}\lt \frac{\pi x}{2}\lt 4$ (that is, when $0\leq x\lt 1$ and when $3\lt x\lt 4$). So the two have the same sign on $(0,1)$, and on $(2,3)$. Lather, rinse, and repeat.

(Of course, this is easier to do using the identity that others have suggested; the approach of looking at the signs of the different factors is what you would use in general, or when there is no easy way of reducing the expression to a single item).


Slighty simpler:$\;$ note $\rm\; ab > 0 \iff a/b > 0\;\;$ so $\;\;\sin(z)\cos(z)>0 \iff \tan(z) > 0$

The rest is trivial using $\;\;\tan(-z) = -\tan(z)\;\;\:$ and $\:\;\;\tan(z+\pi) = \tan(z)$


$$ksin\left(\frac{\pi x}2\right)cos\left(\frac{\pi x}2\right)\gt0\\=>\frac k2sin(\pi x)\gt0$$ Now since k is constant and positive, just ignore k. $sin(\pi x)\gt0$ when $x\in (2x,2x+1)$ where $x \in Z$.


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