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.
 A: $$\begin{pmatrix}
\sin(2A) & \sin C & \sin B \\
\sin C & \sin (2B) & \sin A \\
\sin B & \sin A & \sin (2C) \\
\end{pmatrix} = 
\begin{pmatrix}
\sin(2A) & \sin (\pi-A-B) & \sin (\pi-A-C) \\
\sin (\pi-A-B) & \sin (2B) & \sin (\pi-B-C) \\
\sin (\pi-A-C) & \sin (\pi-B-C) & \sin (2C) \\
\end{pmatrix} $$
$$=
\begin{pmatrix}
\sin A \cos A + \cos A \sin A & \sin A \cos B+\cos A  \sin B & \sin A \cos C + \sin C \cos A \\
\sin B \cos A + \cos B \sin A & \sin B \cos B + \cos B \sin B & \sin B \cos C + \cos B \sin C \\
\sin A \cos C + \sin C \cos A & \sin B \cos C + \cos B \sin C & \sin C \cos C + \cos C \sin C
\end{pmatrix}
$$
$$=\begin{pmatrix}
\sin A & \cos A \\ \sin B & \cos B \\ \sin C & \cos C 
\end{pmatrix} 
\begin{pmatrix} 
\cos A & \cos B & \cos C \\ \sin A & \sin B & \sin C 
\end{pmatrix}$$
My other post has a moral: exponentials are easier than trig functions. I can't see a moral in this one.
A: Put $u = e^{i a}$, $v = e^{i b}$ and $w = e^{i c}$. Note that $uvw = e^{i(a+b+c)} = -1$.
After multiplying all entries by $2i$, we are looking at
$$\begin{pmatrix} 
u^2-u^{-2} & v-v^{-1} & w-w^{-1} \\ 
v-v^{-1} & w^2-w^{-2} & u-u^{-1} \\
w-w^{-1} & u-u^{-1} & v^2-v^{-2} \\
\end{pmatrix} = $$
$$\begin{pmatrix}
-u^{-2} & v & w \\
v & -w^{-2} & u \\
w & u & -v^{-2} \\
\end{pmatrix} - 
\begin{pmatrix}
-u^2 & v^{-1} & w^{-1} \\
v^{-1} & -w^2 & u^{-1} \\
w^{-1} & u^{-1} & -v^2 \\
\end{pmatrix} =
 $$
$$-\begin{pmatrix}
u^{-2} & u^{-1} w^{-1} & u^{-1} v^{-1} \\
u^{-1}w^{-1} & w^{-2} & v^{-1} w^{-1} \\
u^{-1} v^{-1} & v^{-1} w^{-1} & v^{-2} \\
\end{pmatrix} +
\begin{pmatrix}
u^2 & uw & uv \\
uw & w^2 & vw \\
uv & uw & v^2 \\
\end{pmatrix}.$$
These two matrices are clearly rank $1$, so their difference is rank $\leq 2$. 
I had no luck finding a geometric interpretation. The kernel of the matrix is $(\sin(c-b), \sin(b-a), \sin(a-c))$, in case that inspires someone else.
Until greater insight comes along, I will take this as a demonstration that trig identites are much easier in exponential form.
A: Here is a proof using a computer algebra system, which does unfortunately involve expanding the determinant. Without loss of generality, suppose that $C$ is the largest angle, so $A,B \leq \pi/2$, hence we can write
$$
\begin{align*}
\cos A &= \sqrt{1-\sin^2 A} \\
\cos B &= \sqrt{1-\sin^2 B} \\
\sin 2A &= 2\sin A\cos A \\
\sin 2B &= 2\sin B\cos B \\
\sin C &= \sin A \cos B + \sin B \cos A \\
\cos C &= \sin A \sin B - \cos A \cos B \\
\sin 2C &= 2\sin C \cos C
\end{align*}
$$
Now we can symbolically calculate the determinant (in the variables $\sin A,\sin B$), and it equals zero.
If you want something smarter, I suggest you try to find a linear dependency on the rows.
A: 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.
