I am trying to prove an equation is always positive given certain constraints for the variables, the equation I've come up with is as follows
$ 2 b \cos(t+2r) \sin(a t) - 2 \sin(t) \cos(a t + 2 b (\pi - r - t)) + 2 \sin(t) \cos(a t) - 2 b \cos(t) \sin(a t)$
In my previous post it is commented that this expression can be simplified to the following equation
$ \sin(b(\pi-r-t)) \sin(t) \sin(b(\pi - r - t) + a t) - b \sin(r) \sin(a t) \sin(r+t)$
Is this correct and if so, how did they do it? Like I said, I'm not a mathematician so I haven't been able to recreate it but if this is true it could prove useful.
Pretty sure the RHS is zero right? The coefficient seems to be canceled out.
\begin{align} & 2 b \cos(t+2r) \sin(a t) - 2 \sin(t) \cos(a t + 2 b (\pi - r - t)) + 2 \sin(t) \cos(a t) - 2 b \cos(t) \sin(a t)\\ &= 2 b [\cos(t+2r)- \cos(t)] \sin(a t) -2[\cos(a t)-\cos(a t + 2 b (\pi - r - t))]\sin(t)\\ &= -4b\sin(r)\sin(r+t)\sin(at) -4 \sin(b (-r - t + π)) \sin(a t + b (-r - t + π))\sin(t)\\ &=-4\left[b\sin(r)\sin(r+t)\sin(at) + \sin(b (-r - t + π)) \sin(a t + b (-r - t + π))\sin(t)\right] \end{align}
Discarding the $(-4)$, you get your expression. This is just using a standard product/sum formula.