If A=cos2π33sin2π33-sin2π33cos2π33, then A2017

A=cos2π33sin2π33-sin2π33cos2π33 let θ=2π33 ∴ A2=cos θsin θ-sin θcos θ cos θsin θ-sin θcos θ =cos2θsin2θ-sin2θcos2θSimilarly, A2017=A2013.A4 =cos (2013θ)sin (2013θ)-sin (2013θ)cos (2013θ).A4 =cos4026 π33sin4026 π33-sin4026 π33cos4026 π33 A4 =cos (122 π)sin (122 π)-sin (122 π)cos (122 π) A4 =1001 A4 A2017=A4

If A=cos2π33sin2π33-sin2π33cos2π33, then A2017

A=cos2π33sin2π33-sin2π33cos2π33 let θ=2π33 ∴ A2=cos θsin θ-sin θcos θ cos θsin θ-sin θcos θ =cos2θsin2θ-sin2θcos2θSimilarly, A2017=A2013.A4 =cos (2013θ)sin (2013θ)-sin (2013θ)cos (2013θ).A4 =cos4026 π33sin4026 π33-sin4026 π33cos4026 π33 A4 =cos (122 π)sin (122 π)-sin (122 π)cos (122 π) A4 =1001 A4 A2017=A4

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If $A = [\begin{matrix} \cos \frac{2 π}{33} & \sin \frac{2 π}{33} \\ - \sin \frac{2 π}{33} & \cos \frac{2 π}{33} \end{matrix}], then A^{2017}$

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$A^{4}$

$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$A^{2}$

answer is C.

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Detailed Solution

$A = [\begin{matrix} \cos \frac{2 π}{33} & \sin \frac{2 π}{33} \\ - \sin \frac{2 π}{33} & \cos \frac{2 π}{33} \end{matrix}] let θ = \frac{2 π}{33} ∴ A^{2} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] = [\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{matrix}]$

$Similarly, A^{2017} = A^{2013} . A^{4} = [\begin{matrix} \cos (2013 θ) & \sin (2013 θ) \\ - \sin (2013 θ) & \cos (2013 θ) \end{matrix}] . A^{4} = [\begin{matrix} \cos (\frac{4026 π}{33}) & \sin (\frac{4026 π}{33}) \\ - \sin (\frac{4026 π}{33}) & \cos (\frac{4026 π}{33}) \end{matrix}] A^{4} = [\begin{matrix} \cos (122 π) & \sin (122 π) \\ - \sin (122 π) & \cos (122 π) \end{matrix}] A^{4} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] A^{4} A^{2017} = A^{4}$