Question

If $\theta \varepsilon R$, then the determinant $\Delta= \begin{vmatrix} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin (\theta+2 \pi / 3) & \cos (\theta+2 \pi / 3) & \sin (2 \theta+4 \pi / 3) \\ \sin (\theta-2 \pi / 3) & \cos (\theta-2 \pi / 3) & \sin (2 \theta-4 \pi / 3) \end{vmatrix}$ equals

• A. $-\sin \theta-\cos \theta$
• B. $\sin 2 \theta$
• C. $1+\sin 2 \theta-\cos 2 \theta$
• D. 0

Answer 1

Answer: (D)

Applying $R_2 \rightarrow R_2+R_3$, and using
$\sin (A+B)+\sin (A-B)=2 \sin A \cos B$ and
$\cos (A+B)+\cos (A-B)=2 \cos A \cos B$, we get
$\begin{vmatrix}\sin \theta&\cos \theta&\sin 2 \theta\\ 2 \sin \theta \cos \left(2\pi/3\right)&2 \cos \theta \cos \left(2\pi/3\right)&2 \sin 2\theta \cos\left(4\pi/3\right) \\ \sin \left(\theta-2\pi/3\right)&\cos\left(\theta-2\pi/3\right)&\sin\left(2\theta-4\pi/3\right)\end{vmatrix}$
$\Delta=\begin{vmatrix}\sin \theta & \cos \theta & \sin 2 \theta \\ -\sin \theta & -\cos \theta & -\sin 2 \theta \\ \sin (\theta-2 \pi / 3) & \cos (\theta-2 \pi / 3) & \sin (2 \theta-4 \pi / 3)\end{vmatrix}$
$\text { since } \cos (2 \pi / 3)=\cos (\pi-\pi / 3)=-\cos (\pi / 3)=-1 / 2 $
$\text { and } \cos (4 \pi / 3)=\cos (\pi+\pi / 3)=-\cos (\pi / 3)=-1 / 2$
$\Rightarrow \Delta=0 \left[\because R_1 \text { and } R_2 \text { are proportional. }\right]$