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According to my calculator,

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 \\ \sin 4 & \sin 5 & \sin 6 \\ \sin 7 & \sin 8 & \sin 9\end{bmatrix}=0$

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 & \sin 4 \\ \sin 5 & \sin 6 & \sin 7 & \sin 8 \\ \sin 9 & \sin 10 & \sin 11 & \sin 12 \\ \sin 13 & \sin 14 & \sin 15 & \sin 16 \end{bmatrix}=0$

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 & \sin 4 & \sin 5 \\ \sin 6 & \sin 7 & \sin 8 & \sin 9 & \sin 10 \\ \sin 11 & \sin 12 & \sin 13 & \sin 14 & \sin 15 \\ \sin 16 & \sin 17 & \sin 18 & \sin 19 & \sin 20 \\ \sin 21 & \sin 22 & \sin 23 & \sin 24 & \sin 25\end{bmatrix}=0$

I conjecture that, for $n\ge 3$,

$\det \begin{bmatrix} \sin 1 & \sin 2 & \sin 3 & \dots & \sin n \\ \sin (n+1) & \sin (n+2) & \sin (n+3) & \dots & \sin (2n) \\ \sin (2n+1) & \sin (2n+2) & \sin (2n+3) & \dots & \sin(3n) \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \sin ((n-1)n+1) & \sin ((n-1)n+2) & \sin ((n-1)n+3) & \dots & \sin (n^2) \end{bmatrix}=0$

Is my conjecture true?


I have only been able to prove the case with $n=3$.

$\sin 5 + \sin 7 + [\sin 1 + \sin (-1)] + [\sin 3 + \sin (-3)]$
$=\sin 5 + \sin 7 + [\sin 1 + \sin (-1)] + [\sin 3 + \sin (-3)]$

Rearrange each side:

$[\sin (-3) + \sin 5] + [\sin 1 + \sin 3] + [\sin (-1) + \sin 7]$
$=[\sin (-1) + \sin 3] + [\sin (-3) + \sin 7] + [\sin 1 + \sin 5]$

Use the product-to-sum formulas:

$(\sin 1)(\cos 4) + (\sin 2)(\cos 1) + (\sin 3)(\cos 4)$
$=(\sin 1)(\cos 2) + (\sin 2)(\cos 5) + (\sin 3)(\cos 2)$

Subtract $(\sin 1)(\cos 14)+(\sin 2)(\cos 13)+(\sin 3)(\cos 12)$ from both sides:

$(\sin 1)(\cos 4 - \cos 14) + (\sin 2)(\cos 1 - \cos 13) + (\sin 3)(\cos 4 - \cos 12)$
$=(\sin 1)(\cos 2 - \cos 14) + (\sin 2)(\cos 5 - \cos 13) + (\sin 3)(\cos 2 - \cos 12)$

Use the product-to-sum formulas again:

$(\sin 1)(\sin 5)(\sin 9)+(\sin 2)(\sin 6)(\sin 7)+(\sin 3)(\sin 4)(\sin 8)$ $=(\sin 1)(\sin 6)(\sin 8)+(\sin 2)(\sin 4)(\sin 9)+(\sin 3)(\sin 5)(\sin 7)$

which is equivalent to

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 \\ \sin 4 & \sin 5 & \sin 6 \\ \sin 7 & \sin 8 & \sin 9\end{bmatrix}=0$

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    $\begingroup$ And idea would be to use the trigonometric formula $\sin(a+1)=\sin a \cos 1+ \cos a \sin 1$ to transform the last column and try to prove that the columns are linearly dependent vectors. $\endgroup$ Sep 9, 2023 at 7:54
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    $\begingroup$ Using this opportunity to once more promote the Pearl Dive. I generally find displays of ad hoc, out-of-the-box thinking the spice of the site. We have too many answerers simply turning the crank. If you see something special, do bring it up in the Pearl Dive for others to evaluate and willing sponsors to take a look. $\endgroup$ Sep 11, 2023 at 8:24
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    $\begingroup$ Related: math.hawaii.edu/home/pdf/putnam/Putnam_2009.pdf $\endgroup$
    – Integrand
    Sep 25, 2023 at 19:45
  • $\begingroup$ Not the same, but similar: this post is related $\endgroup$ Feb 1 at 22:49

5 Answers 5

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Every one of the functions $\sin(x), \ldots, \sin(x+n)$ solves the ODE $y'' + y = 0$. But the solution space of a linear (homogeneous) second order ODE is only two-dimensional, which tells you that these sine functions must be linearly dependent if you take at least $3$ of them. That is, there exist $\lambda_1,\ldots,\lambda_n \in \mathbb{C}$ not all $0$ such that $$\sum\limits_{j = 1}^n \lambda_j \sin(x+j) = 0$$ for all $x$. Now plug in $x = 0$, $x = n$ and so on. This shows that your columns are linearly dependent for $n \geq 3$ and hence the determinant must be $0$.

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By applying the Sum-to-product formulas we get that

$\sin\!\big(2n+i\big)+\sin i=2\sin\!\big(n+i\big)\cos n$

for any $\,i\in\big\{1,2,\cdots,n\big\}\,.$

Consequently ,

$\sin\!\big(2n+i\big)=\color{red}{-1}\!\cdot\!\sin i+\color{red}{2\cos n}\!\cdot\!\sin\!\big(n+i\big)$

for any $\,i\in\big\{1,2,\cdots,n\big\}\,.$

It means that the third row of the matrix $M$ is a linear combination of the first two rows.

So the rows of the matrix $M$ form a linearly dependent set, hence the determinant of $M$ is equal to $\,0\,.$

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No computation is necessary. The entries of $M$ are given by \begin{eqnarray*} M(k,l) = \sin(1+nk+\ell) = \sin(1+nk)\cos\ell + \cos(1+nk)\sin\ell. \end{eqnarray*} This formula shows that $M$ is the sum of two matrices with rank $1$, so rank$(M) \le 2$. Hence $M$ is not invertible if $n \ge 3$.

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Hint

Use the formula $\sin(a+2)= \sin a \cos 2 + \cos a \sin 2$ in the last column, the formula $\sin(a+1) = \sin a \cos 1 + \cos a \sin 1$ in the penultimate column and develop the determinant. You’ll get a linear combination of 4 determinants all having two identical columns if $n \ge 3$. Hence the given determinant vanishes.

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Using complex matrices. Indeed, let $N\in \mathbb C^{n\times n}$ such that $N_{k,\ell} = e^{i(n(k-1)+\ell)}$ it is clear that the $\ell$ column of $N$ is $e^{i\ell}$ times the first column. So $N$ is rank $1$ matrix. Since $M={\mathrm{ Im}}(N)=(N-\overline{N})/2i$. You have $\mathrm{rank}(M) \le \mathrm{rank}(N) + \mathrm{rank}(\overline N)= 2$ so $\det(M)=0$ if $n\ge 3$

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  • $\begingroup$ Why my answer was down voted? $\endgroup$
    – Kroki
    Feb 1 at 14:54

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