Question

The value of
$\left|\begin{array}{ccc}sin\theta & cos\theta & sin2\theta \\ sin(\theta +\frac{2\pi }{3})& cos(\theta +\frac{2\pi }{3})& sin(2\theta +\frac{4\pi }{3})\\ sin(\theta -\frac{2\pi }{3})& cos(\theta -\frac{2\pi }{3})& sin(2\theta -\frac{4\pi }{3})\end{array}\right|$ is

• A. zero for few values of $\theta$.
• B. zero for all values of $\theta$.
• C. zero for no value of $\theta$.
• D. zero for one value of $\theta$.

Answer 1

The correct option is B zero for all values of $\theta$.

Let $\Delta =\left|\begin{array}{ccc}sin\theta & cos\theta & sin2\theta \\ sin(\theta +\frac{2\pi }{3})& cos(\theta +\frac{2\pi }{3})& sin(2\theta +\frac{4\pi }{3})\\ sin(\theta -\frac{2\pi }{3})& cos(\theta -\frac{2\pi }{3})& sin(2\theta -\frac{4\pi }{3})\end{array}\right|$
Applying $R_2\to R_2+R_3$
$=\left|\begin{array}{ccc}sin\theta & cos\theta & sin2\theta \\ sin(\theta +\frac{2\pi }{3})+sin(\theta -\frac{2\pi }{3})& cos(\theta +\frac{2\pi }{3})+cos(\theta -\frac{2\pi }{3})& sin(2\theta +\frac{4\pi }{3})+sin(2\theta -\frac{4\pi }{3})\\ sin(\theta -\frac{2\pi }{3})& cos(\theta -\frac{2\pi }{3})& sin(2\theta -\frac{4\pi }{3})\end{array}\right|$
$sin(\theta +\frac{2\pi }{3})+sin(\theta -\frac{2\pi }{3})=2sin\theta cos\frac{2\pi }{3}$
$=2sin\theta cos(\pi -\frac{\pi }{3})$
$=-2sin\theta cos\frac{\pi }{3}=-sin\theta$
and $cos(\theta +\frac{2\pi }{3})+cos(\theta +\frac{2\pi }{3})$
$=2cos\theta cos\left(\frac{2\pi }{3}\right)=2cos\theta (-\frac{1}{2})=-cos\theta$
and $sin(2\theta +\frac{4\pi }{3})+cos(2\theta -\frac{4\pi }{3})$
$=2sin2\theta cos\frac{4\pi }{3}=2sin2\theta cos(\pi +\frac{\pi }{3})$
$=-2sin2\theta cos\frac{\pi }{3}=-sin2\theta$
$\therefore \Delta =\left|\begin{array}{ccc}sin\theta & cos\theta & sin2\theta \\ -sin\theta & -cos\theta & -sin2\theta \\ sin(\theta -\frac{2\pi }{3})& cos(\theta -\frac{2\pi }{3})& sin(2\theta -\frac{4\pi }{3})\end{array}\right|=0$
[Since, $R_1$ and $R_2$ are proportional]