<s>[INST] A triangle determinant that is always zero How do we prove, without actually expanding, that
$$\begin{vmatrix} 
\sin {2A}& \sin {C}& \sin {B}\\
\sin{C}& \sin{2B}& \sin {A}\\
\sin{B}& \sin{A}& \sin{2C}
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried adding and subtracting from the rows and columns and I even tried using the sine rule, but to no avail.
 [/INST] Using the Law of Sines, we can write
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $a$, $b$, $c$ are the sides, and $d$ is the circumdiameter, of the triangle. And the Law of Cosines gives us
$$\cos A = \frac{1}{2bc}(-a^2+b^2+c^2) \qquad\text{, etc.}$$
With $\sin 2 x = 2 \sin x \cos x$, we can express the determinant as
$$\left|\begin{array}{ccc}
\frac{a}{d}\frac{-a^2+b^2+c^2}{bc} & \frac{c}{d} & \frac{b}{d} \\[4pt]
\frac{c}{d} & \frac{b}{d}\frac{a^2-b^2+c^2}{ca} & \frac{a}{d} \\[4pt]
\frac{b}{d} & \frac{a}{d} & \frac{c}{d}\frac{a^2+b^2-c^2}{ab}
\end{array}\right|$$
From here, we can "factor-out" $\frac{1}{dbc}$, $\frac{1}{dca}$, $\frac{1}{dab}$ from the first, second, and third rows:
$$\frac{1}{dbc}\frac{1}{dca}\frac{1}{dab}\;\left|\begin{array}{ccc}
a(-a^2+b^2+c^2) & b c^2 & c b^2 \\[4pt]
a c^2 & b (a^2-b^2+c^2) & c a^2 \\[4pt]
ab^2 & b a^2 & c (a^2+b^2-c^2)
\end{array}\right|$$
Then, we factor-out $a$, $b$, $c$ from first, second, and third columns:
$$\frac{a b c}{d^3a^2b^2c^2}\;\left|\begin{array}{ccc}
-a^2+b^2+c^2 & c^2 & b^2 \\[4pt]
c^2 & a^2-b^2+c^2 & a^2 \\[4pt]
b^2 & a^2 & a^2+b^2-c^2
\end{array}\right|$$
Subtracting, say, the first row from the second and third gives
$$\frac{1}{d^3abc}\;\left|\begin{array}{ccc}
-a^2+b^2+c^2 & c^2 & b^2 \\[4pt]
a^2-b^2 & a^2-b^2 & a^2-b^2 \\[4pt]
a^2-c^2 & a^2-c^2 & a^2-c^2
\end{array}\right|$$
which clearly vanishes.
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