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 Post subject: A determinant involving a trianglePosted: Wed May 18, 2016 9:02 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
Let a triangle have angles $A, B,C$. Evaluate the determinant:

$$\mathscr{D}=\begin{vmatrix} 1 &\sin A & \cot \frac{A}{2} \\\\ 1& \sin B & \cot \frac{B}{2}\\\\ 1& \sin C & \cot \frac{C}{2} \end{vmatrix}$$

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: A determinant involving a trianglePosted: Sat Oct 22, 2016 9:54 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
We give a solution.

\begin{align*}
\mathcal{D} &=\begin{vmatrix}
1 & \sin A &\cot \frac{A}{2} \\\\
0 &\sin B - \sin A &\cot \frac{B}{2}- \cot \frac{A}{2} \\\\
0& \sin C - \sin B & \cot \frac{C}{2} - \cot \frac{B}{2}
\end{vmatrix} \\\\
&= \begin{vmatrix}
\sin B - \sin A & \cot \frac{B}{2}- \cot \frac{A}{2} \\\\
\sin C - \sin A & \cot \frac{C}{2} - \cot \frac{A}{2}
\end{vmatrix}\\\\
&= \begin{vmatrix}
2 \sin \frac{B-A}{2} \cos \frac{B+A}{2} &\frac{\sin \frac{B-A}{2}}{\sin \frac{B}{2} \sin \frac{A}{2}} \\\\
2 \sin \frac{C-A}{2} \cos \frac{C+A}{2} & \frac{\sin \frac{C-A}{2}}{\sin \frac{C}{2} \sin \frac{A}{2}}
\end{vmatrix}\\\\
&= 2 \sin \frac{B-A}{2} \sin \frac{C-A}{2} \begin{vmatrix}
\sin \frac{C}{2} &\frac{1}{\sin \frac{B}{2} \sin \frac{A}{2}} \\\\
\sin \frac{B}{2}& \frac{1}{\sin \frac{C}{2} \sin \frac{A}{2}}
\end{vmatrix}\\
&= 2 \sin \frac{B-A}{2} \sin \frac{C-A}{2} \frac{\sin \frac{C}{2} \sin \frac{B}{2} - \sin \frac{B}{2} \sin \frac{C}{2}}{\sin \frac{A}{2} \sin \frac{B}{2}\sin \frac{C}{2}} \\
&=0
\end{align*}

It is worth reading Mollweide's formula.

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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