# Using polar coordinates to find solutions to equation

Using polar coodinates, find all the points $$(x,y)\ne 0$$ such that $$\frac{2xy}{x^2+y^2}=\frac{1}{2}$$.

I thought I could somehow substitute $$x=r\cos(\theta), y=r\sin(\theta)$$ but I'm not sure how to proceed. What strategy should I use to tackle these types of problems? Thanks!

EDIT: The numerator should be $$2xy$$

• Why didn't you try ? – Yves Daoust Feb 17 at 8:59

$$x = r\cos\theta, y = r\sin\theta$$

$$\frac{2xy}{x^2+y^2} = \frac{2r^2\sin\theta\cos\theta}{r^2} = \sin 2\theta$$

Solve $$\sin 2\theta = \frac 12$$ in the first four quadrants (only two quadrants are applicable).

Then, to get the solution in terms of $$x$$ and $$y$$, think about the equation(s) of line(s) subtending that angle with the horizontal axis. What is the gradient of those line(s)?

• I have $(r\cos(5\pi/12),rsin(5\pi/12))$ and $(r\cos(\pi/12),rsin(\pi/12)) \forall x\in \mathbb{R}$. Is this correct? – Sam Kirkiles Feb 17 at 8:57
• I would probably express it as $(x, x\tan\frac{\pi}{12}) \cup (x,x\tan\frac{5\pi}{12}) \ \forall x \in \mathbb{R} \backslash {0}$ to eliminate the dummy variable $r$. If you want to make your life a bit harder, you can explicitly work out the value of the trigonometric ratios in terms of irrational numbers, but I don't think that's necessary. – Deepak Feb 17 at 9:40

The substitution is exactly what you should use, except that you don't have $$x=r\cos (x)$$, but rather $$x=r\cos(\phi)$$.

Then, using the fact that $$\sin^2\phi + \cos^2\phi = 1$$, the denominator on your right side should simplify quite a bit.

As $$(x,y)\neq0$$, you have $$x=r\cos(\theta)$$ and $$y=r\sin(\theta)$$ for $$r>0$$ and $$\theta \in [0,2\pi]$$.

$$\frac{2xy}{x^2+y^2}=\frac{2r^2 \cos(\theta)\sin(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))} = 2\cos(\theta)\sin(\theta)=\sin(2\theta)$$
So you get to solve $$\sin(2\theta) = \frac{1}{2}$$
By your work $$r^2=4r^2\cos\theta,$$ which gives $$r=0,$$ which is impossible, or $$1=4\cos\theta\sin\theta,$$ which is $$\sin2\theta=\frac{1}{2}.$$