Matrices And Determinants Ques 6
- Prove that for all values of $\theta$
$ \left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin \left(\theta+\frac{2 \pi}{3}\right) & \cos \left(\theta+\frac{2 \pi}{3}\right) & \sin \left(2 \theta+\frac{4 \pi}{3}\right) \\ \sin \left(\theta-\frac{2 \pi}{3}\right) & \cos \left(\theta-\frac{2 \pi}{3}\right) & \sin \left(2 \theta-\frac{4 \pi}{3}\right) \end{array}\right|=0 $
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Solution:
Let $\Delta=\left|\begin{array}{ccc}\sin \theta & \cos \theta & \sin 2 \theta \\ \sin \left(\theta+\frac{2 \pi}{3}\right) & \cos \left(\theta+\frac{2 \pi}{3}\right) & \sin \left(2 \theta+\frac{4 \pi}{3}\right) \\ \sin \left(\theta-\frac{2 \pi}{3}\right) & \cos \left(\theta-\frac{2 \pi}{3}\right) & \sin \left(2 \theta-\frac{4 \pi}{3}\right)\end{array}\right|$
Applying $R_2 \rightarrow R_2+R_1$
$ =\left|\begin{array}{lll} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin \left(\theta+\frac{2 \pi}{3}\right) & \cos \left(\theta+\frac{2 \pi}{3}\right) & \sin \left(2 \theta+\frac{4 \pi}{3}\right) \\ +\sin \left(\theta-\frac{2 \pi}{3}\right) & +\cos \left(\theta-\frac{2 \pi}{3}\right) & +\sin \left(2 \theta-\frac{4 \pi}{3}\right) \\ \sin \left(\theta-\frac{2 \pi}{3}\right) & \cos \left(\theta-\frac{2 \pi}{3}\right) & \sin \left(2 \theta-\frac{4 \pi}{3}\right) \end{array}\right| $
Now, $\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta-\frac{2 \pi}{3}\right)$
$ \begin{aligned} & =2 \sin \left(\frac{\theta+\frac{2 \pi}{3}+\theta-\frac{2 \pi}{3}}{2}\right) \cos \left(\frac{\theta+\frac{2 \pi}{3}-\theta+\frac{2 \pi}{3}}{2}\right) \\ & =2 \sin \theta \cos \frac{2 \pi}{3}=2 \sin \theta \cos \left(\pi-\frac{\pi}{3}\right) \\ & =-2 \sin \theta \cos \frac{\pi}{3}=-\sin \theta \\ & \text { and } \cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta-\frac{2 \pi}{3}\right) \\ & =2 \cos \left(\frac{\theta+\frac{2 \pi}{3}+\theta-\frac{2 \pi}{3}}{2}\right) \cos \left(\frac{\theta+\frac{2 \pi}{3}-\theta+\frac{2 \pi}{3}}{2}\right) \\ & =2 \cos \theta \cos \left(\frac{2 \pi}{3}\right)=2 \cos \theta\left(-\frac{1}{2}\right)=-\cos \theta \\ & \text { and } \sin \left(2 \theta+\frac{4 \pi}{3}\right)+\sin \left(2 \theta-\frac{4 \pi}{3}\right) \\ & =2 \sin \left(\frac{2 \theta+\frac{4 \pi}{3}+2 \theta-\frac{4 \pi}{3}}{2}\right) \cos \left(\frac{2 \theta+\frac{4 \pi}{3}-2 \theta+\frac{4 \pi}{3}}{2}\right) \\ & =2 \sin 2 \theta \cos \frac{4 \pi}{3}=2 \sin 2 \theta \cos \left(\pi+\frac{\pi}{3}\right) \\ & =-2 \sin 2 \theta \cos \frac{\pi}{3}=-\sin 2 \theta \\ & \therefore \Delta=\left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\ -\sin \theta & -\cos \theta & -\sin 2 \theta \\ \sin \left(\theta-\frac{2 \pi}{3}\right) & \cos \left(\theta-\frac{2 \pi}{3}\right) & \sin \left(2 \theta-\frac{4 \pi}{3}\right) \end{array}\right|=0 \\ \end{aligned} $
[since, $R_1$ and $R_2$ are proportional]
BETA
