Problem source

\textit{In this question, do not consider the special cases in which the denominators of any of your expressions are zero.}
Express $\tan(\theta_1+\theta_2+\theta_3+\theta_4)$ in terms of $t_i$, where  $t_1=\tan\theta_1\,$, etc.
Given that $\tan\theta_1$, $\tan\theta_2$, $\tan\theta_3$ and $\tan\theta_4$ are the four roots of the equation \[at^4+bt^3+ct^2+dt+e=0
\] (where $a\ne0$), find an expression in terms
of $a$, $b$, $c$,  $d$ and $e$ for $\tan(\theta_1+\theta_2+\theta_3+\theta_4)$.
The four real numbers  $\theta_1$, $\theta_2$, $\theta_3$ and $\theta_4$ lie in the range $0\le \theta_i<2\pi$ and satisfy the equation 
\[
p\cos2\theta+\cos(\theta-\alpha)+p=0\,,\]
where $p$ and  $\alpha$ are independent of $\theta$. Show that $\theta_1+\theta_2+\theta_3+\theta_4=n\pi$ for some integer $n$.

Solution source

\begin{align*}
\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &= \frac{\tan(\theta_1 + \theta_2) + \tan(\theta_3 + \theta_4)}{1 - \tan(\theta_1 +\theta_2)\tan(\theta_3+\theta_4)} \\
&= \frac{\frac{t_1+t_2}{1-t_1t_2}+\frac{t_3+t_4}{1-t_3t_4}}{1-\frac{t_1+t_2}{1-t_1t_2}\frac{t_3+t_4}{1-t_3t_4}} \\
&= \frac{(t_1+t_2)(1-t_3t_4)+(t_3+t_4)(1-t_1t_2)}{(1-t_1t_2)(1-t_3t_4)-(t_1+t_2)(t_3+t_4)} \\
&= \frac{t_1 +t_2+t_3+t_4 - (t_1t_2t_3+t_1t_2t_4+t_1t_3t_4+t_2t_3t_4)}{1-t_1t_2-t_1t_3-t_1t_4-t_2t_3-t_2t_4-t_3t_4}
\end{align*}

If $t_1, t_2, t_3, t_4$ are the roots of $at^4+bt^3+ct^2+dt+e = 0$, then $t_1+t_2+t_3+t_4 = -\frac{b}{a}, t_1t_2+t_1t_3+t_1t_4+t_2t_3+t_2t_4+t_3t_4 = \frac{c}{a}, t_1t_2t_3+t_1t_2t_4+t_1t_3t_4+t_2t_3t_4 = -\frac{d}{a}$, therefore the expression is:

\begin{align*}
\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &= \frac{-\frac{b}{a}+\frac{d}{a}}{1 - \frac{c}{a}} \\
&= \frac{d-b}{a-c}
\end{align*}

\begin{align*}
&&0 &= p \cos 2\theta + \cos (\theta - \alpha) + p \\
&&&= p (2\cos^2 \theta -1) + \cos \theta \cos \alpha - \sin \theta \sin \alpha + p \\
&&&= 2p \cos^2 \theta + \cos \theta \cos \alpha - \sin \theta \sin \alpha\\
\Rightarrow && 0 &=2p \cos \theta + \cos \alpha - \tan \theta \sin \alpha \\
\Rightarrow && -2p \cos \theta&=  \cos \alpha - \tan \theta \sin \alpha \\
\Rightarrow && 4p^2 \cos^2 \theta &= \cos^2 \alpha - 2 \sin \alpha \cos \alpha \tan \theta + \sin^2 \alpha \tan^2 \theta  \\
&& 4p^2 \frac{1}{1 + \tan^2 \theta} &= \cos^2 \alpha - \sin 2\alpha \tan \theta + \sin^2 \alpha \tan^2 \theta \\
\Rightarrow && 4p^2 &= \cos^2 \alpha - \sin 2\alpha t+t^2-\sin2\alpha t^3+\sin^2 \alpha t^4 \\
\Rightarrow && \tan (\theta_1+\theta_2 + \theta_3+ \theta_4) &= \frac{0}{\sin^2 \alpha - 1} \\
&&&= 0 \\
\Rightarrow && \theta_1 + \theta_2 + \theta_3 + \theta_4 &= n\pi
\end{align*}