I want to prove that $f(\psi)$ defined below, has at least one root. I tried using intermediate value theorem but I could not prove the existence of a root. I would appreciate it if anybody could help.
$f(\psi)=\sum_{i=1}^K \frac{\alpha_i sin(\psi+\theta_i)}{2+cos(\psi-\theta_i)}$
$\alpha_i$ and $\theta_i$ are given for every $i$ and we have
$\forall i \in \{1,2,...,K\} : 0 \leq \alpha_i \leq 1$ and $-\pi \leq \theta_i \leq \pi$.
Assume that $f$ has no roots. "It can be shown" that $f$ is continuous (critically, the denominator is never equal to $0$), so $f$ is therefore everywhere greater (or less) than $0$. Without loss of generality, assume
$$\forall \psi\ f(\psi) > 0$$
Therefore, the integral of $f$ with respect to $\psi$ is always greater than $0$, and moreover, always increasing. Considering just the interval $[0,2\pi]$, then, we have
$$\int_{0}^{2\pi} f(\psi)\ d\psi \neq 0$$
Since $f$ is a sum, its integral is equal to the sum of its summands' integrals on the same interval.
$$\int_{0}^{2\pi} \sum_{i} f_i(\psi)\ d\psi = \sum_{i} \int_{0}^{2\pi}f_i(\psi)\ d\psi$$
Each of these summands $f_i$ is a periodic function centered at $0$:
$$f_i(\psi)=\frac{\alpha_i sin(\psi+\theta_i)}{2+cos(\psi-\theta_i)}$$
"It can be shown" that over the period of $2\pi$ each $f_i$ spends equal amounts of time above and below $0$. That is,
$$\int_{0}^{2\pi}f_i(\psi)\ d\psi = 0$$
$$\therefore \sum_{i} \int_{0}^{2\pi}f_i(\psi)\ d\psi = 0$$
$$\therefore \int_{0}^{2\pi} f(\psi)\ d\psi = 0$$
But this is a contradiction: this integral is supposed to be positive. Therefore, our initial assumption is false: $f$ must have a root.
It's a long comment and because of this I write it here. If $0 \leq \theta_i \leq \pi$ then $0\leq\sin(\theta_i)\leq 1$ and because of $2+\cos(\theta_i)>0$ and $2-\cos(\theta_i)>0$, we have:
\begin{align} 0\leq f(0)=\sum_{i=1}^K \frac{\alpha_i sin(\theta_i)}{2+cos(\theta_i)} < \sum_{i=1}^K \alpha_i sin(\theta_i)< K\\ \\ 0\geq f(\pi)=\sum_{i=1}^K \frac{-\alpha_i sin(\theta_i)}{2-cos(\theta_i)} > \sum_{i=1}^K -\alpha_i sin(\theta_i)> -K \end{align} then there is a $\psi\in [0,\pi]$, such that $f(\psi)=0$.
If $-\pi \leq \theta_i \leq 0$ then $-1\leq\sin(\theta_i)\leq 0$ and because of $2+\cos(\theta_i)>0$ and $2-\cos(\theta_i)>0$, we have:
\begin{align} 0\geq f(0)=\sum_{i=1}^K \frac{\alpha_i sin(\theta_i)}{2+cos(\theta_i)} > \sum_{i=1}^K \alpha_i sin(\theta_i)> -K\\ \\ 0\leq f(-\pi)=\sum_{i=1}^K \frac{-\alpha_i sin(\theta_i)}{2-cos(\theta_i)} < \sum_{i=1}^K -\alpha_i sin(\theta_i)< K \end{align} then there is a $\psi\in [-\pi,0]$, such that $f(\psi)=0$.
Now, if $-\pi \leq \theta_i \leq \pi$, then we can rearrange the function $f(\psi)$ as follows: $$f(\psi)=g(\psi)+h(\psi)$$ such that $0 \leq \theta_i(g(\psi)) \leq \pi$ and $-\pi \leq \theta_i(h(\psi)) \leq 0$. Therefore, there is a $\psi_1\in [0,\pi]$ where $g(\psi_1)=0$ and there is a $\psi_2\in [-\pi,0]$ where $h(\psi_2)=0$.
My conjecture: There is a $\psi_3\in [\psi_2,\psi_1]$ such that $f(\psi_3)=0$.
I hope it be useful.
