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So the problem is to find all points $(x,y)$ on the real plane such that $f(x,y) = \cos^2(x+t) + 2\sin(x+t)\cos(y) - \frac{(\cos y - 1)^2}{2} - \sin(x) \lt .5$ for all real $t$. I'm not sure where to start with this, I don't think I fully understand the equation... how can the result be in terms of $f(x,y)$ when $t$ is also essentially a variable? shouldn't this be a 3 dimensional plot?

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2 Answers 2

My instinct says you want to get rid of that $t$. Maybe you could find the value of $t$ that makes $f(x,y,t)$ maximum?

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To help understand the question, let's make the two $t$'s into different variables. Now we want $f(x,y) = \cos^2(x+t_1) + 2\sin(x+t_2)\cos(y) - \frac{(\cos y - 1)^2}{2} - \sin(x) \lt .5$ for all real $t_1,t_2$. For any $x$, we can find $t_1$ to make the $\cos^2$ anything from $0$ to $1$ and we can find $t_2$ to make the $\sin$ term anything from $-1$ to $1$. If we want the RHS to be $\lt .5$ for all $t_1,t_2$, we can set the $\cos^2(x+t_1)$ term to $1$ (as that is worst case) and the $\sin(x+t_2)$ term to $\pm 1$ to match the sign of $\cos y$. Now we demand that $f(x,y)=1+2|\cos y| - \frac{(\cos y - 1)^2}{2} - \sin(x) \lt .5$ or $2f(x,y)-1=-\cos^2 y + \cos y +2|\cos y|-\sin x \lt 0$. You can set $u=\sin x, v=\cos y$, with $u,v \in [-1,1]$ and see that you need $-v^2+v+2|v|-u\lt 0$, find the region in the $u,v$ unit square that satisfies this, and translate it back to the $x,y$ plane.

The image I got of this function is below. u is vertical, v is horizontal. The allowable region is the part away from the -u axis

graph with separate $t$'s

Your problem is more difficult because $t_1$ and $t_2$ are correlated, in fact the same. All the region that satisfied the uncorrelated version will satisfy yours, but there may be others.

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