If in a triangle ABC, sin3A + sin3B + sin3C = 3sinAsinBsinc, then what is the value of the determinant \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b\end{vmatrix}\) where a,b,c are sides of the triangle?
To solve this problem, we need to establish a connection between the given trigonometric identity and the properties of the determinant. We are provided with the trigonometric condition involving angles of a triangle \(A\), \(B\), and \(C\) as follows:
\(\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C\)
Recall that for a triangle with angles \(A\), \(B\), and \(C\), we have:
\(A + B + C = 180^\circ\)
The given identity is a form of the well-known identity for angles in a triangle:
\(\sin^3 A + \sin^3 B + \sin^3 C - 3 \sin A \sin B \sin C = 0\)
which simplifies to:
\((\sin A + \sin B + \sin C)(\sin^2 A + \sin^2 B + \sin^2 C - \sin A \sin B - \sin B \sin C - \sin C \sin A) = 0\)
The non-trivial solution here is:
\(\sin A + \sin B + \sin C = 0\)which occurs when one of the angles is \(180^\circ\), an invalid configuration for a triangle.
Thus, we proceed with the determinant. The matrix presented is:
| \(a\) | \(b\) | \(c\) |
| \(b\) | \(c\) | \(a\) |
| \(c\) | \(a\) | \(b\) |
The determinant of this circulant matrix is:
\(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}\)
This determinant is known to be zero when:
1. Sides \(a\), \(b\), and \(c\) do not all follow the conditions needed for a non-degenerate matrix determinant by being unequal and not aligning.
Hence, the value of the determinant is:
0
This rules out other options based on the non-degenerate nature of a triangle and the relationships shown by the sines of angles conditions and supports the correct choice which turns out to be:
0
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