Prove determinant of a matrix with trigonometry functions.

Question

I am a student. I am 15. Actually, I am a new learner of matrix & determinant. I have to prove an equation. I tried my best to solve this. But, I am failed.

The equation is:

$$ \begin{vmatrix} 1 & \cos2\alpha & \sin\alpha \\ 1 & \cos2\beta& \sin\beta \\ 1 & \cos2\gamma & \sin\gamma \\ \end{vmatrix} = 2(\sin\alpha - \sin\beta)(\sin\beta - \sin\gamma)(\sin\gamma - \sin\alpha) $$

Thanks-

Answer 1

First idea: Make $0$'s in the first column. For example: $Row_2-Row_1$ and $Row_3-Row_1$

$$\text{Your det}=\begin{vmatrix} 1&\cos2\alpha&\sin\alpha\\ 0&\cos2\beta-\cos2\alpha&\sin\beta-\sin\alpha\\ 0&\cos2\gamma-\cos2\alpha&\sin\gamma-\sin\alpha\end{vmatrix} = \begin{vmatrix} \cos2\beta-\cos2\alpha&\sin\beta-\sin\alpha\\\cos2\gamma-\cos2\alpha&\sin\gamma-\sin\alpha\end{vmatrix}$$

Now, as @Ethan said in comments, try to use the double angle formula to continue.

Answer 2

By the rule of SARRUS we get $$\cos(2\beta)\sin(\gamma)+\cos(2\alpha)\sin(\beta)+\cos(2\gamma)\sin(\alpha)-\cos(2\alpha)\sin(\gamma)-\cos(2\gamma)\sin(\beta)-\cos(2\beta)\sin(\alpha)$$