Matrices and Determinants 2 Question 31
35. Prove that for all values of $\theta$
(\left|| \sin \theta\right.) | $\cos \theta$ | $\sin 2 \theta$ | | :—: | :—: | :—: | | $\sin \theta+\frac{2 \pi}{3}$ | $\cos \theta+\frac{2 \pi}{3}$ | $\sin 2 \theta+\frac{4 \pi}{3}$ | | $\sin \theta-\frac{2 \pi}{3}$ | $\cos \theta-\frac{2 \pi}{3}$ | $\sin 2 \theta-\frac{4 \pi}{3}$ |$|=0$
(2000, 3M)
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Solution:
- Let $\Delta=$
$$ \begin{aligned} & \sin \theta+\frac{2 \pi}{3} \quad \cos \theta+\frac{2 \pi}{3} \quad \sin 2 \theta+\frac{4 \pi}{3} \\ & \sin \theta-\frac{2 \pi}{3} \quad \cos \theta-\frac{2 \pi}{3} \quad \sin 2 \theta-\frac{4 \pi}{3} \end{aligned} $$
Applying $R _2 \rightarrow R _2+R _3$
$$ =\mid \begin{array}{lll} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin \theta+\frac{2 \pi}{3} & \cos \theta+\frac{2 \pi}{3} & \sin 2 \theta+\frac{4 \pi}{3} \\ +\sin \theta-\frac{2 \pi}{3} & +\cos \theta-\frac{2 \pi}{3} & +\sin 2 \theta-\frac{4 \pi}{3} \\ \sin \theta-\frac{2 \pi}{3} & \cos \theta-\frac{2 \pi}{3} & \sin 2 \theta-\frac{4 \pi}{3} \end{array} $$
Now, $\sin \theta+\frac{2 \pi}{3}+\sin \theta-\frac{2 \pi}{3}$
$$ \begin{aligned} & =2 \sin \frac{\theta+\frac{2 \pi}{3}+\theta-\frac{2 \pi}{3}}{2} \cos \frac{\theta+\frac{2 \pi}{3}-\theta+\frac{2 \pi}{3}}{2} \\ & =2 \sin \theta \cos \frac{2 \pi}{3}=2 \sin \theta \cos \pi-\frac{\pi}{3} \\ & =-2 \sin \theta \cos \frac{\pi}{3}=-\sin \theta \end{aligned} $$
and $\cos \theta+\frac{2 \pi}{3}+\cos \theta-\frac{2 \pi}{3}$
$$ \begin{aligned} & =2 \cos \frac{\theta+\frac{2 \pi}{3}+\theta-\frac{2 \pi}{3}}{2} \cos \frac{\theta+\frac{2 \pi}{3}-\theta+\frac{2 \pi}{3}}{2} \\ & =2 \cos \theta \cos \frac{2 \pi}{3}=2 \cos \theta-\frac{1}{2}=-\cos \theta \end{aligned} $$
and $\sin 2 \theta+\frac{4 \pi}{3}+\sin 2 \theta-\frac{4 \pi}{3}$
$$ \begin{aligned} & =2 \sin \frac{2 \theta+\frac{4 \pi}{3}+2 \theta-\frac{4 \pi}{3}}{2} \cos \frac{2 \theta+\frac{4 \pi}{3}-2 \theta+\frac{4 \pi}{3}}{2} \\ & =2 \sin 2 \theta \cos \frac{4 \pi}{3}=2 \sin 2 \theta \cos \pi+\frac{\pi}{3} \\ & =-2 \sin 2 \theta \cos \frac{\pi}{3}=-\sin 2 \theta \\ & \therefore \Delta=\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\ -\sin \theta & -\cos \theta & -\sin 2 \theta \\ \sin \theta-\frac{2 \pi}{3} & \cos \theta-\frac{2 \pi}{3} & \sin 2 \theta-\frac{4 \pi}{3} \end{array}=0 \end{aligned} $$
[since, $R _1$ and $R _2$ are proportional]
