# Perplexities on the Weierstrass substituition for $\phi=\pi$

Let us suppose that we have a system of equations including trigonometric expressions in $\phi$ and we want to bound the number of possible solutions.

If I apply the Weierstrass substituition $t=\tan(\phi/2)$ I come up with a system in $t$ . If $\phi=\pi$ was a solution, there will be no corresponding t (only $t -> \infty$).

The number of possible solutions can be [the degree of the polynomium in t] + 1 (i.e. $\phi=\pi$).

For example

$sin(\phi)=0$

$\frac{2t}{1+t^2}=0$

and we get only t=0 (we see that $\lim_{t->\infty}=0$ as well)

As a workaround, I use the substitution $\phi=\phi'+\epsilon$,$t'=tan(\phi'/2)$ and leave $\epsilon$ as a parameter, so that I can "rotate" all the solutions (if they are finite) in a way that none falls in $\pi$.

Is there a more elegant solution?

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Let $A\sin x+B\cos x=C--->(1)$

Applying the well-known formula $A=R\cos \theta,B=R\sin \theta,$ where $R>0$

$A^2+B^2=R^2$ and $\tan \theta=\frac BA$

So, $A\sin x+B\cos x=\sqrt{A^2+B^2}\cos(x-\theta)$

For the real values of $x,-1\le \cos(x-\theta)\le 1\implies \sqrt{A^2+B^2}\le C\le \sqrt{A^2+B^2}$,

if $C$ outside this range, there will b no solution of in $R$.

If $\sqrt{A^2+B^2}\le C\le \sqrt{A^2+B^2}, \cos(x-\theta)=\frac C{\sqrt{A^2+B^2}}=\cos \alpha$(say)

So, $x-\theta=2n\pi\pm \alpha\implies x=2n\pi+ \theta\pm \alpha$ where $n$ is any integer.

Clearly, there is exactly one solution in $r\pi<x\le (r+1)\pi$ where $r$ is any integer.

Now, if we put $t=\tan \frac x2,$ from $(1)$

we get, $A\frac{2t}{1+t^2}+B\frac{1-t^2}{1+t^2}=C$ or $(C+B)t^2-2At+C-B=0$

This is a quadratic equation in $t,$ so there are two roots, which correspond to the one root mentioned above.

If the coefficient of $t^2$ is zero, i.e., $C+B=0$ there is one infinite root, the other is the root of $-2At+C-B=0$.

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