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Sin(A+B)Sin(A-B)

sinθ cosθ t...

Question

$∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ sin (θ + \frac{2 π}{3}) & cos (θ + \frac{2 π}{3}) & sin (2 θ + \frac{4 π}{3}) sin (θ - \frac{2 π}{3}) & cos (θ - \frac{2 π}{3}) & sin (2 θ - \frac{4 π}{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$ equals

A

3 sin θ

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B

{sin}^{3} θ

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C

0

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D

None of these

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Solution

The correct option is A None of these
$∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ sin (θ + \frac{2 π}{3}) & cos (θ + \frac{2 π}{3}) & sin (2 θ + \frac{4 π}{3}) sin (θ - \frac{2 π}{3}) & cos (θ - \frac{2 π}{3}) & sin (2 θ - \frac{4 π}{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

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

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ sin (θ + \frac{2 π}{3}) + sin (θ - \frac{2 π}{3}) & cos (θ + \frac{2 π}{3}) + cos (θ - \frac{2 π}{3}) & sin (2 θ + \frac{4 π}{3}) + sin (2 θ - \frac{4 π}{3}) sin (θ - \frac{2 π}{3}) & cos (θ - \frac{2 π}{3}) & sin (2 θ - \frac{4 π}{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

We know that $sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2})$

and $cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2})$

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ 2 sin θ cos \frac{2 π}{3} & 2 cos θ cos \frac{2 π}{3} & 2 sin 2 θ cos \frac{4 π}{3} sin (θ - \frac{2 π}{3}) & cos (θ - \frac{2 π}{3}) & sin (2 θ - \frac{4 π}{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

We know that $sin (- θ) = - sin θ$ and $cos (- θ) = cos θ$

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ 2 sin θ cos \frac{2 π}{3} & 2 cos θ cos \frac{2 π}{3} & 2 sin 2 θ cos \frac{4 π}{3} - sin (\frac{2 π}{3} - θ) & cos (\frac{2 π}{3} - θ) & - sin (\frac{4 π}{3} - 2 θ) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ 2 sin θ cos (π - \frac{π}{3}) & 2 cos θ cos (π - \frac{π}{3}) & 2 sin 2 θ cos (π + \frac{π}{3}) - sin (π - \frac{π}{3} - θ) & cos (π - \frac{π}{3} - θ) & - sin (π + \frac{π}{3} - 2 θ) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$
We know that $sin (π - θ) = sin θ, cos (π - θ) = - cos θ, sin (π + θ) = - sin θ, cos (π + θ) = cos θ$

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ - 2 sin θ cos \frac{π}{3} & - 2 cos θ cos \frac{π}{3} & 2 sin 2 θ cos \frac{π}{3} - sin (π - (\frac{π}{3} + θ)) & cos (π - (\frac{π}{3} + θ)) & - sin (π + (\frac{π}{3} - 2 θ)) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ - 2 sin θ cos \frac{π}{3} & - 2 cos θ cos \frac{π}{3} & 2 sin 2 θ cos \frac{π}{3} - sin (\frac{π}{3} + θ) & - cos (\frac{π}{3} + θ) & sin (\frac{π}{3} - 2 θ) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

We know that $cos \frac{π}{3} = \frac{1}{2}$

$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ - 2 sin θ \times \frac{1}{2} & - 2 cos θ \times \frac{1}{2} & 2 sin 2 θ \times \frac{1}{2} - sin (\frac{π}{3} + θ) & - cos (\frac{π}{3} + θ) & sin (\frac{π}{3} - 2 θ) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin θ & cos θ & tan θ - sin θ & - cos θ & sin 2 θ - sin (\frac{π}{3} + θ) & - cos (\frac{π}{3} + θ) & sin (\frac{π}{3} - 2 θ) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

$R_{1} \to R_{1} + R_{2}$

$= ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & tan θ + sin 2 θ - sin θ & - cos θ & sin 2 θ - sin (\frac{π}{3} + θ) & - cos (\frac{π}{3} + θ) & sin (\frac{π}{3} - 2 θ) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

$= 0 - 0 + (tan θ + sin 2 θ) (sin θ cos (\frac{π}{3} + θ) - cos θ sin (\frac{π}{3} + θ))$

We know that $sin (A - B) = sin A cos B - cos A sin B$

$= (tan θ + sin 2 θ) sin (θ - \frac{π}{3} - θ)$

$= - (tan θ + sin 2 θ) sin \frac{π}{3}$

$= - \frac{\sqrt{3}}{2} (tan θ + sin 2 θ)$

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