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I'll state the question from my book below:

If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of $$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$

Here's how I tried solving the problem:

$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$

$R_2 \to R_2 - R_1$

$R_3 \to R_3 -R_1$

$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$

Expanding the determinant along $C_3$

\begin{align} &= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\ &= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\ &= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\ &= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\ &= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\ &= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C} \end{align}

I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.

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$$F=\begin{vmatrix} \sin^2B-\sin^2A & \cot B-\cot A \\ \sin^2C-\sin^2A & \cot C-\cot A \end{vmatrix}$$

$$=\begin{vmatrix} \sin^2B-\sin^2A & -\dfrac{\sin(B-A)}{\sin A\sin B} \\ \sin^2C-\sin^2A & -\dfrac{\sin(C-A)}{\sin C\sin A} \end{vmatrix}$$

$$=\dfrac1{\sin B\sin^2A\sin C}\begin{vmatrix}\sin(B-A)\sin(B+A)\sin B\sin A&-\sin(B-A)\\ \sin(C-A)\sin(C+A)\sin C\sin A&-\sin(C-A)\end{vmatrix}$$

Using $A+B+C=\pi,\sin(B+A)=\sin C$ etc.,

$$F=\dfrac{\sin A\sin B\sin C}{\sin^2A\sin B\sin C}\begin{vmatrix}\sin(B-A)&-\sin(B-A)\\\sin(C-A)&-\sin(C-A)\end{vmatrix}=?$$

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  • $\begingroup$ Thanks for the answer. I understood everything other than how you got to the expression $=\dfrac1{\cos B\cos^2A\cos C}\begin{vmatrix}\sin(B-A)\sin(B+A)\sin B\sin A&-\sin(B-A)\\ \sin(C-A)\sin(C+A)\sin C\sin A&-\sin(C-A)\end{vmatrix}$ from the previous one. $\endgroup$ – SamInuyasha ANMF Feb 26 '18 at 13:07
  • $\begingroup$ @SamInuyashaANMF, Please find the rectified answer $\endgroup$ – lab bhattacharjee Feb 26 '18 at 16:01
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In the standard notation we obtain: $$\Delta=\sum_{cyc}\sin^2\alpha(\cot\beta-\cot\gamma)=\sum_{cyc}\frac{4S^2}{b^2c^2}\left(\frac{\frac{a^2+c^2-b^2}{2ac}}{\frac{2S}{ac}}-\frac{\frac{a^2+b^2-c^2}{2ab}}{\frac{2S}{ab}}\right)=$$ $$=S\sum_{cyc}\frac{a^2+c^2-b^2-a^2-b^2+c^2}{b^2c^2}=2S\sum_{cyc}\frac{c^2-b^2}{b^2c^2}=2S\sum_{cyc}\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=0.$$

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  • $\begingroup$ Thanks for the answer. I'm sorry, I'm probably not aware of the standard notation you are talking of. But how do you get $\sin A = \frac {2S} {bc}$? Which formula did you use? $\endgroup$ – SamInuyasha ANMF Feb 26 '18 at 13:15
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By Euler's theorem $O,G,H$ are collinear. By considering their trilinear coordinates it follows that $$ \det\begin{pmatrix}\cos(A) & \cos(B) & \cos(C) \\ \frac{1}{\sin A}&\frac{1}{\sin B}&\frac{1}{\sin C}\\ \frac{1}{\cos A}&\frac{1}{\cos B}&\frac{1}{\cos C}\end{pmatrix}=0$$ and by multiplying the first column by $\cos(A)$, the second column by $\cos(B)$ and the third column by $\cos(C)$ we get that $$ \det\begin{pmatrix}\cos^2(A) & \cos^2(B) & \cos^2(C) \\ \cot(A)&\cot(B)&\cot(C)\\ 1&1&1\end{pmatrix}=0$$ and by replacing the first row with the difference between the third row and the first row $$ \det\begin{pmatrix}\sin^2(A) & \sin^2(B) & \sin^2(C) \\ \cot(A)&\cot(B)&\cot(C)\\ 1&1&1\end{pmatrix}=0$$ readily follows. One may also notice that if $\theta\in\{A,B,C\}$ then $$ 8R^2\cdot \sin^2\theta + 4[ABC]\cdot\cot\theta -(a^2+b^2+c^2)\cdot 1 = 0 $$ hence the nullspace of the given matrix has dimension $\geq 1$.

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A hint, as requested:

What you've done looks pretty good; I haven't checked every bit of the algebra, but the symmetry of the result makes me think that you've probably done it right.

What you have not done is use the important fact given at the start: that $A, B,$ and $C$ are the angles of a triangle (hence they sum to $\pi$). I don't see right away how to use that fact, but it's clearly important.

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  • $\begingroup$ Umm, John, isn’t $\cot(\pi/2)=0$? $\endgroup$ – Lubin Feb 25 '18 at 14:45
  • $\begingroup$ Ooops. More coffee needed. I'll edit. :) $\endgroup$ – John Hughes Feb 25 '18 at 15:10
  • $\begingroup$ I used that fact as well when I got $sin (A+B) = \sin C$ and $sin (A+C) = sin B$. $\endgroup$ – SamInuyasha ANMF Feb 26 '18 at 12:37
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You solved it all right , just took cosA.cos C in denominator instead of sinA.SinC

$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$

$R_2 \to R_2 - R_1$

$R_3 \to R_3 -R_1$

$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$

Expanding the determinant along $C_3$

$= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A)$

$= \sin(B+A) \sin(B-A)[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}] - [\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}]\sin(C+A) \sin(C-A)$

$= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\sinA \sinc C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\sinA \sin C}$

$= \frac {\sin(B-A) \sin (A-C)} {\sin A} [\frac {\sin(A+B)} {\sin C} - \frac {\sin(A+C)} {\sin B}]$

$= \frac {\sin(B-A) \sin (A-C)} {\sin A} [\frac {\sin C} {\sin C} - \frac {\sin B} {\sin B}]$

$= \frac {\sin(B-A) \sin (A-C) \sin (A)} [\{0}]$

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