Prove that this determinant is $0$

Question

$$ \begin{vmatrix} \cos 2x & \cos 2y & \cos 2u \\ \sin^2 x & \sin^2 y & \sin^2 u \\ 1 & 1 & 1 \end{vmatrix} = 0 $$

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.

Answer 1

Here's how easy it is if you do use $\cos(2x) = 1-2\sin^2(x)$:

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

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.

Answer 2

By using the equality $cos(2x) = 2cos^2(x)-1 =1-2sin^2(x)$, -2 second line + third line = first line

Answer 3

Via a cofactor expansion along the top row, you have $$\begin{align*}\begin{vmatrix}\cos2x&\cos2y&\cos2u\\\sin^2x&\sin^2y&\sin^2u\\1&1&1\end{vmatrix}&=\cos2x\begin{vmatrix}\sin^2y&\sin^2u\\1&1\end{vmatrix}-\cos2y\begin{vmatrix}\sin^2x&\sin^2u\\1&1\end{vmatrix}+\cos2u\begin{vmatrix}\sin^2x&\sin^2y\\1&1\end{vmatrix}\end{align*}$$ Since $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$, this is equivalent to $$\cos2x(\sin^2y-\sin^2u)-\cos2y(\sin^2x-\sin^2u)+\cos2u(\sin^2x-\sin^2y)$$ Assuming you know the half-angle identity for sine, i.e. $\dfrac{1-\cos2x}{2}=\sin^2x$, you can write this as $$\frac{\cos2x(\cos2u-\cos2y)-\cos2y(\cos2u-\cos2x)+\cos2u(\cos2y-\cos2x)}{2}$$ Notice the disappearing terms.

Answer 4

$$\text{ Given determinant }=-\left|\begin{matrix}\sin^2x-\cos^2x&\sin^2y-\cos^2y&\sin^2u-\cos^2u\\\sin^2x&\sin^2y&\sin^2u\\1&1&1\end{matrix}\right|$$ $$=-\left|\begin{matrix}\sin^2x&\sin^2y&\sin^2u\\\sin^2x&\sin^2y&\sin^2u\\1&1&1\end{matrix}\right| \text{ , by using } R_1'=R_1-R_2+R_3 .$$ $$=0$$

Answer 5

If you replace $\cos$ by an arbitrary function $g$, the determinant (e.g. from expanding across the top row) is $$ g(2x) (\sin^2 y - \sin^2 u) + g(2y) (\sin^2 u - \sin^2 x) + g(2u) (\sin^2 x - \sin^2 y)$$ and this will not be $0$ in general. Take any $x,y,u$ such that no two of $\sin^2 x$, $\sin^2 y$ and $\sin^2 u$ are equal, and you have $a g(2x) + b g(2y) + c g(2z)$ for some nonzero $a,b,c$, and there's no reason for this to be $0$. So without some trig identity that can relate $g$ to $\sin$, it's impossible to do this problem.