# 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.

## 4 Answers

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.

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.

$$\begin{pmatrix} \sin(2A) & \sin C & \sin B \\ \sin C & \sin (2B) & \sin A \\ \sin B & \sin A & \sin (2C) \\ \end{pmatrix} = \begin{pmatrix} \sin(2A) & \sin (\pi-A-B) & \sin (\pi-A-C) \\ \sin (\pi-A-B) & \sin (2B) & \sin (\pi-B-C) \\ \sin (\pi-A-C) & \sin (\pi-B-C) & \sin (2C) \\ \end{pmatrix}$$

$$= \begin{pmatrix} \sin A \cos A + \cos A \sin A & \sin A \cos B+\cos A \sin B & \sin A \cos C + \sin C \cos A \\ \sin B \cos A + \cos B \sin A & \sin B \cos B + \cos B \sin B & \sin B \cos C + \cos B \sin C \\ \sin A \cos C + \sin C \cos A & \sin B \cos C + \cos B \sin C & \sin C \cos C + \cos C \sin C \end{pmatrix}$$ $$=\begin{pmatrix} \sin A & \cos A \\ \sin B & \cos B \\ \sin C & \cos C \end{pmatrix} \begin{pmatrix} \cos A & \cos B & \cos C \\ \sin A & \sin B & \sin C \end{pmatrix}$$

My other post has a moral: exponentials are easier than trig functions. I can't see a moral in this one.

• But the final result does look rather neat :) – hypergeometric Apr 25 '15 at 15:15

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.