# $sup_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{xy})}-\inf{(\cos{x^2}+\cos{y^2}-\cos{(xy)})}=6$

let $x,y\in R$,prove or disprove $$sup_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{xy})}-\inf_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{(xy)})}=6$$

I think we must show that $$\cos{x^2}+\cos{y^2}-\cos{xy}\in (-3,3)$$ it is clear $$|\cos{x^2}+\cos{y^2}-\cos{xy}|<|\cos{x^2}|+|\cos{y^2}|+|\cos{xy}|<1+1+1=3$$ But I think we must why $$\cos{x^2}+\cos{y^2}-\cos{xy}\to 2.99999\cdots,$$ so I think this problem is key is prove this reslut?

on the other words: why $$\cos{x^2}+\cos{y^2}-\cos{xy}$$ is dense in $(-3,3)$

• Use for $n\in \mathbb{N}$, $x_n=\sqrt{2n\pi}$, $y_n=\sqrt{2(n+1)\pi}$, and the fact that if $n\to +\infty$, we have $x_ny_n=2\pi\sqrt{n^2+n}=(2n+1)\pi+o(1)$ Commented Nov 25, 2014 at 14:10

Let $\displaystyle f(x,y)=\cos(x^2)+\cos(y^2)-\cos(xy)$. It is clear that $-3\leq f(x,y)\leq 3$ for all $x,y$.
a) Let $\displaystyle x_n=\sqrt{2\pi n}$, $\displaystyle y_n=\sqrt{2\pi(n+1)}$. We have $\cos(x_n^2)=\cos(y_n^2)=1$. Put $z_n=x_ny_n$. We have $\displaystyle z_n=2\pi \sqrt{n^2+n}=2\pi n\sqrt{1+\frac{1}{n}}$. As $\displaystyle \sqrt{1+x}=1+\frac{x}{2}+o(x)$, we get $\displaystyle z_n=2\pi n(1+\frac{1}{2n}+o(1/n))=(2n+1)\pi+\varepsilon _n$, with $\varepsilon_n\to 0$. Hence $\cos(z_n)=-\cos(\varepsilon_n)\to -1$, and $f(x_n,y_n)\to 3$. Hence ${\rm Sup}(f(x,y)=3$.
b) Let $\displaystyle a_n=\sqrt{(2n+1)\pi}$, $b_n=\sqrt{(2n+3)\pi}$. We have $\cos(a_n^2)=\cos(b_n^2)=-1$, and $\displaystyle c_n=a_nb_n=\pi(2n+1)\sqrt{1+\frac{2}{2n+1}}$. We have $\displaystyle \sqrt{1+\frac{2}{2n+1}}=1+\frac{1}{2n+1}+o(1/(2n+1))$. Hence $\displaystyle c_n=\pi (2n+1)+\pi+o(1/(2n+1)=2\pi(n+1)+\alpha_n$, with $\alpha_n\to 0$ if $n\to +\infty$. Hence $\cos(a_nb_n)=\cos(\alpha_n)\to 1$, as $n\to +\infty$. We get $f(a_n, b_n)\to -3$ as $n\to \infty$, and ${\rm Inf}\{f(x,y)\}=-3$. So the result $6$ is correct.