Question

If $\theta \epsilon R$, then the determinant $\Delta =\left|\begin{array}{ccc}\sin \theta & \cos \theta & \sin 2\theta \\ \sin (\theta +2\pi /3)& \cos (\theta +2\pi /3)& \sin (2\theta +4\pi /3)\\ \sin (\theta -2\pi /3)& \cos (\theta -2\pi /3)& \sin (2\theta -4\pi /3)\end{array}\right|$ 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 $R_2\to R_2+R_3$, 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
$\left|\begin{array}{ccc}\sin \theta & \cos \theta & \sin 2\theta \\ 2\sin \theta \cos \left(2\pi /3\right)& 2\cos \theta \cos \left(2\pi /3\right)& 2\sin 2\theta \cos \left(4\pi /3\right)\\ \sin \left(\theta -2\pi /3\right)& \cos \left(\theta -2\pi /3\right)& \sin \left(2\theta -4\pi /3\right)\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}\sin \theta & \cos \theta & \sin 2\theta \\ -\sin \theta & -\cos \theta & -\sin 2\theta \\ \sin (\theta -2\pi /3)& \cos (\theta -2\pi /3)& \sin (2\theta -4\pi /3)\end{array}\right|$
$\text{ since }\cos (2\pi /3)=\cos (\pi -\pi /3)=-\cos (\pi /3)=-1/2$
$\text{ and }\cos (4\pi /3)=\cos (\pi +\pi /3)=-\cos (\pi /3)=-1/2$
$\Rightarrow \Delta =0\left[\because R_1\text{ and }{R}_2\text{ are proportional. }\right]$