SATHEE: Matrices and Determinants 2 Question 31

Matrices and Determinants 2 Question 31

35. Prove that for all values of $θ$

(\left|| \sin \theta\right.) | $\cos θ$ | $\sin 2 θ$ | | :—: | :—: | :—: | | $\sin θ + \frac{2 π}{3}$ | $\cos θ + \frac{2 π}{3}$ | $\sin 2 θ + \frac{4 π}{3}$ | | $\sin θ - \frac{2 π}{3}$ | $\cos θ - \frac{2 π}{3}$ | $\sin 2 θ - \frac{4 π}{3}$ | $| = 0$

(2000, 3M)

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Solution:

Let $Δ =$

$\begin{aligned} \sin θ + \frac{2 π}{3} \cos θ + \frac{2 π}{3} \sin 2 θ + \frac{4 π}{3} \\ \sin θ - \frac{2 π}{3} \cos θ - \frac{2 π}{3} \sin 2 θ - \frac{4 π}{3} \end{aligned}$

Applying $R_{2} \to R_{2} + R_{3}$

$=∣ \begin{array}{lll} \sin θ & \cos θ & \sin 2 θ \\ \sin θ + \frac{2 π}{3} & \cos θ + \frac{2 π}{3} & \sin 2 θ + \frac{4 π}{3} \\ + \sin θ - \frac{2 π}{3} & + \cos θ - \frac{2 π}{3} & + \sin 2 θ - \frac{4 π}{3} \\ \sin θ - \frac{2 π}{3} & \cos θ - \frac{2 π}{3} & \sin 2 θ - \frac{4 π}{3} \end{array}$

Now, $\sin θ + \frac{2 π}{3} + \sin θ - \frac{2 π}{3}$

$\begin{aligned} = 2 \sin \frac{θ + \frac{2 π}{3} + θ - \frac{2 π}{3}}{2} \cos \frac{θ + \frac{2 π}{3} - θ + \frac{2 π}{3}}{2} \\ = 2 \sin θ \cos \frac{2 π}{3} = 2 \sin θ \cos π - \frac{π}{3} \\ = - 2 \sin θ \cos \frac{π}{3} = - \sin θ \end{aligned}$

and $\cos θ + \frac{2 π}{3} + \cos θ - \frac{2 π}{3}$

$\begin{aligned} = 2 \cos \frac{θ + \frac{2 π}{3} + θ - \frac{2 π}{3}}{2} \cos \frac{θ + \frac{2 π}{3} - θ + \frac{2 π}{3}}{2} \\ = 2 \cos θ \cos \frac{2 π}{3} = 2 \cos θ - \frac{1}{2} = - \cos θ \end{aligned}$

and $\sin 2 θ + \frac{4 π}{3} + \sin 2 θ - \frac{4 π}{3}$

$\begin{aligned} = 2 \sin \frac{2 θ + \frac{4 π}{3} + 2 θ - \frac{4 π}{3}}{2} \cos \frac{2 θ + \frac{4 π}{3} - 2 θ + \frac{4 π}{3}}{2} \\ = 2 \sin 2 θ \cos \frac{4 π}{3} = 2 \sin 2 θ \cos π + \frac{π}{3} \\ = - 2 \sin 2 θ \cos \frac{π}{3} = - \sin 2 θ \\ ∴ Δ = \begin{array}{ccc} \sin θ & \cos θ & \sin 2 θ \\ - \sin θ & - \cos θ & - \sin 2 θ \\ \sin θ - \frac{2 π}{3} & \cos θ - \frac{2 π}{3} & \sin 2 θ - \frac{4 π}{3} \end{array} = 0 \end{aligned}$