Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:


*

*for all $x$ and $y$  

*for no $x$ and $y$  

*only when $x \neq y \neq 0$  

*only when $x = y \neq 0$ 


Suppose that $x$ and $y$ are not $0$.
We know that $$x^2 + y^2 \geq 2xy. \quad (1)$$
Two cases:


*

*$x>0$ and $y<0$ (or $x<0$ and $y>0$), then $\frac{x^2+y^2}{2xy} < 0$, and you cannot have $\sin^2(\theta) < 0$

*$x>0$ and $y>0$ (or $x<0$ and $y<0$). Then $(1)$ becomes 
$$\frac{x^2+y^2}{2xy} \geq 1.$$
The only solution is to have $\frac{x^2+y^2}{2xy} = 1$.
