A signal in a circuit I am working with can be modeled as
expression (1): $\operatorname{sgn}((\sin(\theta+\phi$))sin($\theta$)
(where $\operatorname{sgn}()$ is the sign function)
I had previously observed that $\sin(\theta)\sin(\theta+\phi)$ is equal to $\frac{1}{2}[\cos(-\phi)-\cos(2\theta+\phi)]$, and that for $\phi=0$ my signal looked like $\lvert\sin(\theta)\rvert$, so I expected my signal could be expressed as some combination of these expressions.
With a bit of plotting and use of excel I've convinced myself that expression(1) is equal to
expression (2): $\operatorname{sgn}(\frac{1}{2}[\cos(-\phi)-\cos(2\theta+\phi)])\lvert\sin(\theta)\rvert$
but I do not have the mathematical prowess to properly show it.
Any assistance provided in showing that expression (1) is equal (or not equal to) expression (2) would be greatly appreciated.
Thanks a lot
\begin{align} \operatorname{sgn}((\sin(\theta+\phi))\sin(\theta) &= \operatorname{sgn}((\sin(\theta+\phi))\operatorname{sgn}(\sin(\theta)|\sin(\theta)|\\ &=\operatorname{sgn}(\sin(\theta+\phi)(\sin(\theta))|\sin(\theta)|\\ &=\operatorname{sgn}(\frac{1}{2}[\cos(-\phi)-\cos(2\theta+\phi)])\lvert\sin(\theta)\rvert \end{align}
because we have $x = \operatorname{sgn}(x)|x|$ (used in the first equality where we apply it on $\sin \theta$) and $\operatorname{sgn}(ab)=\operatorname{sgn}(a)\operatorname{sgn}(b)$ (used in the second equality).