Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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



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?

share|cite|improve this question
up vote 0 down vote accepted

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 $-2At+C-B=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.