how to write the polar form of $x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2$?

Question

$$x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2\\ r=(x^2+y^2)^{1/2}\\ x^2/r^4-y^2/r^4$$

Convert $x^2-y^2$ to polar form $$x=r\cos(\theta)$$ $$y=r\sin(\theta)$$ $$(r\cos(\theta))^2-(r\sin(\theta))^2=r^2 \cos(2 θ)$$

$$r^2 \cos(2 θ)/r^4= \cos(2\theta)/r^2$$

Is that correct?

Answer 1

HINT: it is $$\frac{r^2\cos^2(\theta)-r^2\sin^2(\theta)}{(r^2)^2}$$

Answer 2

Hint

It should look like this in the end:

$$E=x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2=\frac {x^2- y^2}{(x^2 + y^2)^2}=\frac {\cos^2(\theta)- \sin^2(\theta)}{r^2}$$ $$E=\frac {2\cos^2(\theta)- 1}{r^2}=\frac {\cos(2\theta)}{r^2}$$ So it's correct...

Answer 3

Yes, you are correct.

$$x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2\\ \\ x^2/r^4-y^2/r^4 = $$

$$\frac{\cos 2\theta}{r^2}$$