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Previous Year Question (PYQs)
1
If $A,B,C$ are angles of a triangle, then the value of
$
\begin{vmatrix}
\sin 2A & \sin C & \sin B \\
\sin C & \sin 2B & \sin A \\
\sin B & \sin A & \sin 2C
\end{vmatrix}
$
is:
Solution
** Since $A+B+C=\pi$, we have
$\cos A=-\cos(B+C)=\sin B\sin C-\cos B\cos C$ and similarly for $B,C$.
Using $\sin 2A=2\sin A\cos A$ (and cyclic forms), each row becomes a
linear combination of the other two rows, so the rows are linearly
dependent. Hence the determinant is $0$.
$\boxed{0}$
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