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I want to compute this determinant: $$ \begin{vmatrix} \sin(2x)&\sin(3x)&\sin(4x)\\ \sin(3x)&\sin(4x)&\sin(5x)\\ \sin(4x)&\sin(5x)&\sin(6x) \end{vmatrix} $$

SAKLY
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  • I tried to use the formula $\sin(a)\pm\sin(b)=2\sin((a\pm b)/2)\cos ((a\mp b)/2)$ by operating on some columns and rows but no "zero" appears... – SAKLY Feb 17 '15 at 15:34
  • Well, I have tried several methods including your formula but it seems just too tricky to deal with. Are you sure that the result is simple? I just used Maple and it produced a terrible mess.... – Vim Feb 17 '15 at 15:47
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    Hint: $\text{row}_1 + \text{row}_3 = 2\cos(x) \times \text{row}_2$. – achille hui Feb 17 '15 at 15:51

1 Answers1

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Using Euler's formula, your determinant is : $$ D = \begin{vmatrix} \frac{e^{2ix}-e^{-2ix}}{2i} & \frac{e^{3ix}-e^{-3ix}}{2i} & \frac{e^{4ix}-e^{-4ix}}{2i} \\ \frac{e^{3ix}-e^{-3ix}}{2i} & \frac{e^{4ix}-e^{-4ix}}{2i} & \frac{e^{5ix}-e^{-5ix}}{2i} \\ \frac{e^{4ix}-e^{-4ix}}{2i} & \frac{e^{5ix}-e^{-5ix}}{2i} & \frac{e^{6ix}-e^{-6ix}}{2i} \end{vmatrix} $$ Factor out the $2i$ factors, and use multilinearity : you obtain $$ D = \frac1{(2i)^3}\begin{vmatrix} e^{2ix}-e^{-2ix} & e^{3ix}-e^{-3ix} & e^{4ix}-e^{-4ix} \\ e^{3ix}-e^{-3ix} & e^{4ix}-e^{-4ix} & e^{5ix}-e^{-5ix} \\ e^{4ix}-e^{-4ix} & e^{5ix}-e^{-5ix} & e^{6ix}-e^{-6ix} \end{vmatrix}$$ $$= \frac1{(2i)^3}\sum_{(\epsilon_1, \epsilon_2, \epsilon_3)\in\{\pm 1\}^3} \epsilon_1 \epsilon_2 \epsilon_3\begin{vmatrix} e^{2i\epsilon_1 x} & e^{3i\epsilon_2 x} & e^{4i\epsilon_3 x} \\ e^{3i\epsilon_1 x} & e^{4i\epsilon_2 x} & e^{5i\epsilon_3 x} \\ e^{4i\epsilon_1 x} & e^{5i\epsilon_2 x} & e^{6i\epsilon_3 x} \end{vmatrix}$$ Factor out the elements from the first line. Each of the determinant in the sum over $(\epsilon_1, \epsilon_2, \epsilon_3)$ is $$ = e^{(2\epsilon_1 + 3\epsilon_2 + 4\epsilon_3)ix}\ \times\begin{vmatrix} 1 & 1 & 1 \\ e^{i\epsilon_1 x} & e^{i\epsilon_2 x} & e^{i\epsilon_3 x} \\ e^{2i\epsilon_1 x} & e^{2i\epsilon_2 x} & e^{2i\epsilon_3 x} \end{vmatrix}. $$

Now for any $(\epsilon_1, \epsilon_2, \epsilon_3)\in\{\pm 1\}^3$, at least two of the columns will be the same (there's only two choices for each $\epsilon_j$ !) : so the last determinant is always zero and $D=0$.

Sary
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