I've been working on a problem on showing the existence of real linear operators on $\mathbb C$ and have reduced it to the trigonometric equation below. Let $$x,y\in (0,1)$$ $$z,t\in[0,1]$$ $$\alpha \in (-\frac \pi2,\frac \pi2),\alpha\neq0$$ $$\omega \in [-\frac \pi2,\frac \pi2]$$ and $x^2+y^2=z^2+t^2=1$. I've reduced my problem to showing whether or not there exists a $\delta\in \mathbb R$ such that $$\frac{z^2y^2}{x^4}\frac{\sin^2(\omega+\delta)\sin^2(\delta)}{\Big( \cos(\delta)-\frac{\sin(\delta)}{\tan (\alpha)} \Big)\sin^4(\alpha)}+\frac{z^2 \sin^2(\omega+\delta)}{x^2\sin^2(\alpha)} -\frac{t^2 \sin^2(\delta)}{x^2 \sin^2(\alpha)} =\frac{t^2}{y^2}\Big( \cos(\delta)-\frac{\sin(\delta)}{\tan (\alpha)} \Big)$$ i.e. For which values of $x,z,\alpha,\omega$ is there a solution in $\delta$? This equation is quite difficult to solve in terms of $\delta$. I put this into mathematica and it reduced to the equation $$t^2y^2 \frac{\sin(\alpha-\omega)\sin^2(\delta)}{\sin^3(\alpha)}- t^2x^2\Big( \cos(\delta)-\frac{\sin(\delta)}{\tan (\alpha)} \Big)^2-\frac12y^2z^2-y^2z^2\Big( \cos(2\delta)+\cos(2\alpha-\delta)-\cos(\delta)-1 \Big)\frac{\sin^2(\delta+\omega)}{\sin^4(\alpha)}=0$$ However this doesnt seem to simplify it. Is there an easy way to check when solutions to this equation exist?

  • $\begingroup$ I've reduced my problem to ... Maybe post the actual problem, rather than this "reduced" form. $\endgroup$
    – dxiv
    Aug 24, 2017 at 22:12

1 Answer 1


Let $Z=z^2, Y=y^2, X=x^2, T=t^2, Q=cos(\delta)-\frac{\sin(\delta)}{\tan(\alpha)}$

and define $ s(\beta)=\sin^2(\beta)$ then: $$ \ \frac{ZY}{X^2}\frac{s(\omega+\delta)s(\delta)}{Qs^2(\alpha)}+\frac{Zs(\omega+\delta)}{Xs(\alpha)}-\frac{Ts(\delta)}{Xs(\alpha)}=\frac{T}YQ \Longrightarrow \\ \ ZY\cdot s(\omega+\delta)s(\delta)+Xs(\alpha)Z\cdot s(\omega+\delta)Q-Xs(\alpha)T\cdot s(\delta)Q=X^2s^2(\alpha)\frac{T}Y\cdot Q^2 \Longrightarrow \\ \ X^2s^2(\alpha)\frac{T}Y\cdot Q^2-Xs(\alpha)Z\cdot s(\omega+\delta)Q+Xs(\alpha)T\cdot s(\delta)Q-ZY\cdot s(\omega+\delta)s(\delta)=0 \Longrightarrow \\ \ \left(Xs(\alpha)\sqrt{\frac{T}Y}\cdot Q+ T\sqrt{\frac{Y}T}\cdot s(\delta)\right) \left(Xs(\alpha)\sqrt{\frac{T}Y}\cdot Q- Z\sqrt{\frac{Y}T}\cdot s(\omega+\delta)\right) =0 $$

This gives two factors to examine for solutions: $$Xs(\alpha)\sqrt{\frac{T}Y}\cdot Q= - T\sqrt{\frac{Y}T}\cdot s(\delta) \Longrightarrow \\ \boxed{-\frac{Xs(\alpha)}{Y}=\frac{s(\delta)}{Q}} \tag{1} $$$$ Xs(\alpha)\sqrt{\frac{T}Y}\cdot Q= Z\sqrt{\frac{Y}T}\cdot s(\omega+\delta)\Longrightarrow \tag{2} \\ \boxed{\frac{T}Z\frac{Xs(\alpha)}{Y}=\frac{s(\omega+\delta)}{Q}} $$

While this doesn't give a formulation for $\delta$ directly, it is simpler. The factors may also yield some additional insights. For example notice the similarity of the two LHSs, as well as similarity of RHSs. And the first factor does not involve $T$ or $Z$. Perhaps not surprisingly given constraints, $T$ and $Z$ appear related as does $X$ and $Y$.


$\frac{s(\delta)}{Q}$ has a period of $2\pi$. Taking $\alpha=0$ as an example, $\frac{s(\delta)}{Q}$ goes to $\pm\infty$ at $\frac\pi2+\pi k$. If $\alpha\ne 0$ the period is still $2\pi$ but the locations where $\frac{s(\delta)}{Q}$ is infinite move, by up to $\frac\pi2$ (versus $\alpha=0$). Within a given period the value of $\frac{s(\delta)}{Q}$ ranges from $-\infty$ to $\infty$. Each possible (real) value of $\frac{s(\delta)}{Q}$ occurs twice in a period. So for a given value of LHS in (1) there exists a solution: $0 \le \delta < \pi$. This means there is a solution in $\delta$ for all values of all parameters.

  • $\begingroup$ @gene was this answer not helpful? I think it answers the question by showing that solutions exist for all values of the parameters. $\endgroup$
    – Χpẘ
    Sep 10, 2017 at 22:19

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