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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 $$k \sin(\frac {\pi x} {2})\cos(\frac {\pi x} {2})>0$$ where $k$ is some huge sausage which is constant and positive. |
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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 $$ x \in (2m, 2m+1), m \in \mathbb{Z}$$ |
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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). |
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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 $\:\;\;\tan(z+\pi) = \tan(z)$ |
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