Aspire Faculty ID #13736 · Topic: JAMIA MILLIA ISLAMIA MCA 2022 · Just now
JAMIA MILLIA ISLAMIA MCA 2022

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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