So I need to prove that
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
$$
= \begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\
\sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\
\sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\
\end{vmatrix}
$$
Now,
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix} = \begin{vmatrix}
\sin^2(\alpha) & \cos^2(\alpha) – \sin^2(\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos^2(\beta) – \sin^2(\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos^2(\gamma) – \sin^2(\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
Best Answer
$$\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\ \sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\ \sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\ \end{vmatrix}=\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin\alpha\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\alpha \\ \sin(\beta) & \cos(\beta) & \sin\beta\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\beta \\ \sin(\gamma) & \cos(\gamma) & \sin\gamma\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\gamma \\ \end{vmatrix}$$$${}$$
$$=\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin\alpha \\ \sin(\beta) & \cos(\beta) & \sin\beta\\ \sin(\gamma) & \cos(\gamma) & \sin\gamma \\ \end{vmatrix}\color{red}{\cos\delta}+\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \cos\alpha \\ \sin(\beta) & \cos(\beta) & \cos\beta\\ \sin(\gamma) & \cos(\gamma) & \cos\gamma \\ \end{vmatrix}\color{red}{\sin\delta}=0+0=0$$