Prove that this determinant is $0$ $$
\begin{vmatrix}
\cos 2x &amp; \cos 2y &amp; \cos 2u \\
\sin^2 x &amp; \sin^2 y &amp; \sin^2 u \\
1 &amp; 1 &amp; 1
\end{vmatrix} = 0
$$

This is the conclusion where I got from another excercise and I want to prove me it is right. Please don't use the fact that $\cos(2x)=\cos^2x-\sin^2x$. We haven't learned it yet.
Here's how easy it is if you do use $\cos(2x) = 1-2\sin^2(x)$:

$$\begin{align}\left|\begin{matrix} \cos(2x) &amp; \cos(2y) &amp; \cos(2u) \\ \sin^2(x) &amp; \sin^2(y) &amp; \sin^2(u) \\ 1 &amp; 1 &amp; 1\end{matrix} \right| &amp;= \left|\begin{matrix} 1-2\sin^2(x) &amp; 1-2\sin^2(y) &amp; 1-2\sin^2(u) \\ \sin^2(x) &amp; \sin^2(y) &amp; \sin^2(u) \\ 1 &amp; 1 &amp; 1\end{matrix} \right| \\ &amp;= \left|\begin{matrix} 1 &amp; 1 &amp; 1 \\ \sin^2(x) &amp; \sin^2(y) &amp; \sin^2(u) \\ 1 &amp; 1 &amp; 1\end{matrix} \right| - 2\left|\begin{matrix} \sin^2(x) &amp; \sin^2(y) &amp; \sin^2(u) \\ \sin^2(x) &amp; \sin^2(y) &amp; \sin^2(u) \\ 1 &amp; 1 &amp; 1\end{matrix} \right|\end{align}$$

This equals zero because the determinant of any matrix with a repeated row (or column) is $0$.

So maybe it's time to learn some trig.
By using the equality $cos(2x) = 2cos^2(x)-1 =1-2sin^2(x)$, 
-2 second line + third line = first line