<s>[INST] Calculate the determinant of $3\times 3$ matrix with $\sin x$ and powers of $\cos x$ How to calculate the determinant of this matrix
$A=\begin{bmatrix}
\sin x & \cos^2x & 1 \\ 
\sin x & \cos x & 0 \\ 
\sin x & 1 & 1
\end{bmatrix}$
$$\left[A\right]=\begin{vmatrix}
\sin x & \cos^2x & 1 \\ 
\sin x & \cos x & 0 \\ 
\sin x & 1 & 1
\end{vmatrix}=\\=\sin x\begin{vmatrix}
 \cos x& 0\\ 
1 & 1
\end{vmatrix}-\cos^2x\begin{vmatrix}
\sin x & 0 \\ 
\sin x & 1
\end{vmatrix}+\begin{vmatrix}
\sin x & \cos x\\
\sin x & 1\\\end{vmatrix}=\\=\sin x\cos x-\cos^2x\sin x+\sin x-\sin x\cos x =\\=\sin x\left(\cos x-\cos^2x+1-\cos x\right)=\sin x \left(1-\cos ^2x\right)=\\=\sin x\cdot \sin^2x=\sin^3x$$
The path is something like this? I'm using the wrong rule?
 [/INST] $$\left[A\right]=\begin{vmatrix} \sin x & \cos^2x & 1\\ \sin x & \cos x & 0 \\ \sin x & 1 & 1 \end{vmatrix}\begin{vmatrix} \sin x & \cos^2x  \\ \sin x & \cos x  \\ \sin x & 1   \end{vmatrix}=\sin x\cos x+\sin x-\sin x\cos x -\sin x\cos^2x=\\=\sin x\left(1-\cos^2 x\right)=\\=\sin x\cdot \sin^2x=\sin^3 x$$
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