Solving the Determinant for Triangle Sides and Angles

This problem asks us to find the value of a specific determinant involving the sides (\(a, b, c\)) and one angle (\(A\)) of a triangle \(ABC\).

Understanding the Triangle Determinant Problem

We are given a \(3 \times 3\) determinant: \(D = \begin{vmatrix} a^2 & b\sin A & c\sin A \\ b\sin A & 1 & \cos A \\ c\sin A & \cos A & 1 \end{vmatrix}\) Our goal is to simplify this expression using the properties of triangles and basic algebra.

Key Triangle Formulas

To solve this, we need to recall some fundamental formulas related to triangles:

• Sine Rule: For any triangle \(ABC\) with sides \(a, b, c\) opposite to angles \(A, B, C\) respectively, the rule states: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) where \(R\) is the circumradius of the triangle.
• Cosine Rule: This rule relates the sides of a triangle to one of its angles: \(a^2 = b^2 + c^2 - 2bc\cos A\) This formula can be rearranged as: \(b^2 + c^2 - 2bc\cos A = a^2\) This rearranged form will be particularly useful.

Step-by-Step Determinant Calculation

Let's calculate the determinant \(D\) using the cofactor expansion method. We will expand along the first row:

\(D = a^2 \begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} - b\sin A \begin{vmatrix} b\sin A & \cos A \\ c\sin A & 1 \end{vmatrix} + c\sin A \begin{vmatrix} b\sin A & 1 \\ c\sin A & \cos A \end{vmatrix}\)

Now, let's compute the \(2 \times 2\) determinants:

1. \(\begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} = (1 \cdot 1) - (\cos A \cdot \cos A) = 1 - \cos^2 A\)
2. \(\begin{vmatrix} b\sin A & \cos A \\ c\sin A & 1 \end{vmatrix} = (b\sin A \cdot 1) - (\cos A \cdot c\sin A) = b\sin A - c\sin A \cos A\)
3. \(\begin{vmatrix} b\sin A & 1 \\ c\sin A & \cos A \end{vmatrix} = (b\sin A \cdot \cos A) - (1 \cdot c\sin A) = b\sin A \cos A - c\sin A\)

Substitute these back into the expression for \(D\):

\(D = a^2(1 - \cos^2 A) - b\sin A (b\sin A - c\sin A \cos A) + c\sin A (b\sin A \cos A - c\sin A)\)

Now, distribute and simplify:

\(D = a^2\sin^2 A - (b^2\sin^2 A - bc\sin^2 A \cos A) + (bc\sin^2 A \cos A - c^2\sin^2 A)\)

\(D = a^2\sin^2 A - b^2\sin^2 A + bc\sin^2 A \cos A + bc\sin^2 A \cos A - c^2\sin^2 A\)

Combine like terms:

\(D = a^2\sin^2 A - b^2\sin^2 A - c^2\sin^2 A + 2bc\sin^2 A \cos A\)

Factor out \(\sin^2 A\):

\(D = \sin^2 A (a^2 - b^2 - c^2 + 2bc \cos A)\)

Now, let's use the rearranged Cosine Rule: \(b^2 + c^2 - 2bc \cos A = a^2\).

Substitute this into the expression:

\(D = \sin^2 A (a^2 - (b^2 + c^2 - 2bc \cos A))\)

\(D = \sin^2 A (a^2 - a^2)\)

\(D = \sin^2 A (0)\)

\(D = 0\)

Conclusion

By applying the cofactor expansion and utilizing the Cosine Rule for triangles, we find that the value of the given determinant simplifies to zero.