Okay so my adventure starts with this equation
$$|x|+|y|=1$$ This makes a square so what I wanted two do is turn it around so it's sitting "right". My plan was to convert the equation to polar and then add $\pi/4$ to $\theta$.
Conversion to polar
It was a straight forward task and I got this
$$|r\cos\theta| +|r\sin\theta|=1$$Now I needed to clean the equation up a little bit so I collected the $r$ terms as such:
$$\begin{align}1&=|r\cos\theta| +|r\sin\theta|\\&=r|\cos\theta|+r|\sin\theta|\\&=r\left(|\cos\theta|+|\sin\theta|\right)\\1/r&=|\cos\theta|+|\sin\theta|\end{align}$$
Two $\theta$s got messy so I rearranged the equation to have one $\theta$
$$\begin{align}1/r&=|\cos\theta|+|\sin\theta|\\1/r^2&=\cos^2\theta+\sin^2\theta+|2\sin\theta\cos\theta|\\1/r&=\sqrt{1+|\sin2\theta|}\end{align}$$Using the trig identies $\sin^2\theta+\cos^2\theta=1$ and $2\sin\theta\cos\theta=\sin2\theta.$
So now that I have my desired polar equation I went about turning the square. To achieve a $\pi/4$ turn I had to add this to each $\theta$. Which gets me a new "$\theta$". If you recall the sin function within the last equation had the $\theta$ so the new operand in that function becomes
$$2\theta\implies2\theta+\pi/2$$
This is the equation that I desire for a square. It works and I am happy with it. However I need to turn this back into Cartesian coordinates.
Conversion into Cartesian
Now that I have
$$1/r=\sqrt{1+|\sin(2\theta+\pi/2)|}$$
I imideately realise that there are two trig identities hidden in there. First a sin – cosine relation.
$$\sin(2\theta+\pi/2)=\cos(2\theta)$$
then a double angle
$$\cos(2\theta)=2\cos^2\theta-1$$
Now we have
$$1/r=\sqrt{1+|2\cos^2\theta-1|}$$
It is ready to be converted. So I start with $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\left(\frac yx\right)$.
$$\frac1{\sqrt{x^2+y^2}}=\sqrt{1+\left|2\cos^2\left[\arctan\left(\frac yx\right)\right]-1\right|}$$
To see if it is still a square I checked and it is. So I am assuming that the mistake will come up after.Okay so I am using trig identities so evaluate $\cos^2\{\arctan(y/x)\}$
$$\begin{align}
\cos^2\left[\arctan\left(\frac yx\right)\right]
=\frac1{y^2/x^2+1}\end{align}$$
Putting it all together
$$\begin{align}\frac1{\sqrt{x^2+y^2}}&=\sqrt{1+\left|\frac2{y^2/x^2+1}-1\right|}\\\frac1{x^2+y^2}&=1+\left|\frac2{y^2/x^2+1}-1\right|\end{align}$$
This is where I get stuck as I have no idea how to collect the $y$ from the RHS.
I should make it clear that I want a single $y$ term on the LHS and all the $x$ terms on the RHS.
Best Answer
You have $$\frac{1}{x^2+y^2} = 1+\left|\frac{2}{y^2/x^2+1}-1\right| = 1+\frac{|x^2-y^2|}{x^2+y^2},$$ or put differently $$1 = x^2+y^2 + |x^2-y^2|.$$
At this point solving by cases helps. Where $|x|<|y|$ we have $1=2y^2$, and elsewhere $1=2x^2$. Hence we have $$\|(x,y)\|_\infty=\max\{|x|,|y|\} = \frac{1}{\sqrt{2}}$$