I am currently dissecting a paper by Lloyd L. Dines, but i got stuck on the proof of the first proposition because the autor used some fact which I'm not sure I understood, he must have considered it too trivial to explicitly write what he was using. His paper is On the mapping of quadratic forms.
I will expose only the part of his reasoning essential to my doubt, without missing any ideas hopefuly. Given a quadratic form $S:\mathbb{R}^2 \rightarrow \mathbb{R}$ that satisfies $$S(\cos\theta,\sin\theta)=k^2>0, \qquad \text{for} \qquad \theta=-\pi/2, 0, \pi/2$$ Then $S(\cos\theta, \sin\theta)$ can vanish for at most two values of theta between $-\pi/2$ and $\pi/2$, and must be negative for theta between those values, then $S(\cos\theta, \sin\theta)$ must be strictly positive for $\theta \in [-\pi/2,0]$ or $\theta \in [0,\pi/2]$. Wht I didn't understand is why $S$ can only be equal to zero twice for those values. Can anyone show me how to get to this conclusion?
The most general quadratic form can be written as $$S(x,y)=ax^2+bxy+cy^2$$ However according to the assumptions it must satisfy $$S(1,0)=S(0,1)=S(0,-1)=k^2$$ and thus this yields $a=c=k^2$. We can therefore write $$S(\cos\theta, \sin\theta)=k^2(1+\frac{b}{k^2}\sin 2\theta)$$ and from this all claims of the paper follow.
More specifically, $S(\cos\theta, \sin\theta)=0$ is only possible if $\sin2\theta=-\frac{k^2}{b}$, but since $k\neq 0$ this equation has at most two solutions in $(-\pi/2,\pi/2)$. Also if one of the solutions is $\theta=x$, then the other one is given by $\theta=\text{sgn}(-b)\text{sgn}(x)\frac{\pi}{2}-x$.
Furthermore, we note that since $\sin2\theta<0~~ \forall~x\in(-\pi/2,0)$ and positive otherwise we conclude that if $b<0(>0)$ then $S(\cos\theta, \sin\theta)>1>0$ if $x\in(-\pi/2,0)\Big(x\in(0, \pi/2)\Big)$