If $$\Delta = \begin{array}{|ccc|} \sin A & \sin B & \sin C \\ \cos A & \cos B & \cos C \\ \cos^3 A & \cos^3 B & \cos^3 C \\ \end{array} = 0$$ Where A, B, C are the angles of a triangle. What can we say about the triangle? (Is the triangle equilateral, isosceles or scalene)?
My Attempt:
If the triangle is isosceles, then either A=B or B=C or C=A. In this case, two of the rows of the determinant would be equal to one another and hence ∆ would be 0. Hence the triangle would be an isosceles triangle.
The Answer:
The answer key for this question states that we cannot infer anything about the given triangle.
Why is this so? Is the answer key wrong? Clearly, if the triangle is not isosceles and other cases exist, it would be a scalene triangle. Could someone explain the case in which a scalene triangle gives the above result?
\begin{align} \Delta&= \sin\alpha \, \cos\beta \, \cos^3\gamma -\sin\alpha \, \cos\gamma \, \cos^3\beta \\ & -\cos\alpha \, \sin\beta \, \cos^3\gamma +\cos\alpha \, \sin\gamma \, \cos^3\beta \\ &+\cos^3\alpha \, \sin\beta \, \cos\gamma -\cos^3\alpha \, \sin\gamma \, \cos\beta =0 \tag{1}\label{1} . \end{align}
Substitutions \begin{align} \sin\alpha&=\frac a{2R},\quad\dots ,\\ \cos\alpha&=\frac{-a^2+b^2+c^2}{2bc},\quad\dots \end{align}
result in
\begin{align} \Delta=& \frac{(a+b)(b+c)(c+a)}{8\,R\,a^3b^3c^3} \\ &\times (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \\ &\times (a^2-b^2)(a^2-c^2)(b^2-c^2) \\ &= \frac{2S^2}{R\,a^3 b^3 c^3}\cdot(a^2-b^2)(a^2-c^2)(b^2-c^2) =0 \tag{2}\label{2} , \end{align} where $R$ and $S$ are the circumradius and the area of the triangle, respectively.
Obviously, excluding the case of the degenerate triangle for which $S=0$, $\Delta$ can be zero iff at least one pair of the sides has the same length, hence the non-degenerate triangle definitely must be isosceles, which also includes equilateral as a special case.