Given a negative semidefinite matrix $A=\{a_{ij}\}_{i,j\in\{1,2,...,n\}}$, and $\sum_{j=1}^{n}\sin(\theta_{n+1}-\theta_j)=0$. Let
$$B : = \left[\begin{array}{ccccc|c} \tiny\color{red}{-\cos(\theta_{n+1}-\theta_1)}&0&\cdots&0&0&\color{red}{\tiny \cos(\theta_1-\theta_{n+1})}\\ 0&\tiny\color{red}{-\cos(\theta_{n+1}-\theta_2)}&\cdots&0&0&\color{red}{\tiny \cos(\theta_2-\theta_{n+1})}\\ \vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\ 0&0&\cdots&\color{red}{\tiny\color{red}{-\cos(\theta_{n-1}-\theta_{n+1})}}&0&\tiny \color{red}{\cos(\theta_{n-1}-\theta_{n+1})}\\ 0&0&\cdots&0&\tiny \color{red}{-\cos(\theta_{n+1}-\theta_{n})} &\tiny \color{red}{\cos(\theta_{n}-\theta_{n+1})}\\ \hline \tiny \color{red}{\cos(\theta_{n+1}-\theta_1)} &\tiny \color{red}{\cos(\theta_{n+1}-\theta_2)} &\cdots&\tiny \color{red}{\cos(\theta_{n+1}-\theta_{n-1})}&\tiny \color{red}{\cos(\theta_{n+1}-\theta_{n})} &\tiny \color{red}{-\sum_{j=1}^{n}\cos(\theta_{n+1}-\theta_{j})} \end{array}\right]$$
Is $C=A+B$ still negative semidefinite?