I posted a question in such a garbled form that I thought I should repost separately. I have deleted the other post just to remove it from consideration. Apologies for that.
I am playing with a complicated quasi-periodic equation, $f(x)$, which boils down to a summation of sines and cosines. I can't give the full equation because I'm still searching for the right formulation. So for now let's call it
$$f(x)=\sum_{a=0}^b (cos(ax)+sin(ax))$$
Is it possible to produce a formula $g(x)$ that delivers the number of instances, $n$, where $f(x)=0$?
I realise this particular $f(x)$ is trivial, but I'm searching for a solution that will work with much more complicated quasi-periodic functions of the same general form. Or is my only option to write code with a loop counter?
Playing, as usual.
Since $\sin(\pi/4)=\cos(\pi/4) =\frac{\sqrt{2}}{2} =\frac1{\sqrt{2}} $,
$\begin{array}\\ \sin(z)+\cos(z) &=\sqrt{2}(\sin(z)\cos(\pi/4)+\cos(z)\sin(\pi/4))\\ &=\frac1{\sqrt{2}}\sin(z+\pi/4)\\ &=\sqrt{2}(\sin(z)\sin(\pi/4)+\cos(z)\cos(\pi/4))\\ &=-\frac1{\sqrt{2}}\cos(z-\pi/4)\\ \end{array} $
$\begin{array}\\ f(x) &=\sum_{a=0}^b (\cos(ax)+\sin(ax))\\ &=\frac1{\sqrt{2}}\sum_{a=0}^b \sin(ax+\pi/4)\\ &=\frac1{\sqrt{2}}Im(e^{i\pi/4}\sum_{a=0}^b e^{iax})\\ &=\frac1{\sqrt{2}}Im(\frac1{\sqrt{2}}(1+i)\sum_{a=0}^b (e^{ix})^a\\ &=\frac1{2}Im((1+i)\dfrac{1-e^{ix(b+1)}}{1-e^{ix}}) \qquad\text{I could stop here, but I'm feeling masochistic}\\ &=\frac1{2}Im(\dfrac{(1+i)(1-e^{ix(b+1)})(1-e^{-ix})}{(1-e^{ix})(1-e^{-ix})})\\ &=\frac1{2}Im(\dfrac{(1+i)(1-e^{ix(b+1)})(1-e^{-ix})}{1-(e^{ix}+e^{-ix})+1})\\ &=\frac1{2}Im(\dfrac{(1+i)(1-e^{ix(b+1)}-e^{-ix}+e^{ixb})}{2-2\cos(x)})\\ &=\frac1{2}Im(\dfrac{(1+i)(1-\cos(x(b+1)-\cos(-x)+\cos(bx))+i(-\sin(x(b+1))-\sin(-x)+\sin(bx)))}{2-2\cos(x)})\\ &=\dfrac1{2(2-2\cos(x))}Im((1+i)(1-\cos(x(b+1)-\cos(x)+\cos(bx))+i(-\sin(x(b+1))+\sin(x)+\sin(bx))))\\ &=\dfrac1{4(1-\cos(x))}Im((1+i)(1-\cos(x(b+1)-\cos(x)+\cos(bx))\\ &\quad +i(-\sin(x(b+1))+\sin(x)+\sin(bx))))\\ &=\dfrac1{4(1-\cos(x))}Im( (1-\cos(x(b+1)-\cos(x)+\cos(bx))\\ &\quad +i(-\sin(x(b+1))+\sin(x)+\sin(bx)))\\ &\quad+i((1-\cos(x(b+1)-\cos(x)+\cos(bx))\\ &\quad +i(-\sin(x(b+1))+\sin(x)+\sin(bx))))\\ &=\dfrac1{4(1-\cos(x))}Im( (1-\cos(x(b+1)-\cos(x)+\cos(bx))\\ &\quad +i(-\sin(x(b+1))+\sin(x)+\sin(bx)))\\ &\quad+i(1-\cos(x(b+1)-\cos(x)+\cos(bx))\\ &\quad -(-\sin(x(b+1))+\sin(x)+\sin(bx))))\\ &=\dfrac1{4(1-\cos(x))}Im( (1-\cos(x(b+1)-\cos(x)+\cos(bx))\\ &\quad -(-\sin(x(b+1))+\sin(x)+\sin(bx))))\\ &\quad +i(-\sin(x(b+1))+\sin(x)+\sin(bx))\\ &\quad+1-\cos(x(b+1)-\cos(x)+\cos(bx)))\\ &=\dfrac1{4(1-\cos(x))}( -\sin(x(b+1))+\sin(x)+\sin(bx)\\ &\quad +1-\cos(x(b+1)-\cos(x)+\cos(bx)))\\ &=\dfrac1{4(1-\cos(x))}( 1 -(\sin(x(b+1))+\cos(x(b+1))+(\sin(x)-\cos(x))+\sin(bx)+\cos(bx)))\\ &=\dfrac1{4(1-\cos(x))}( 1 -\sqrt{2}(\sin(x(b+1)+\pi/4)+(\sin(x-\pi/4))+\sin(bx+\pi/4)))\\ \end{array} $
I figure that this has a 10% chance of being correct.