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Question

$$\displaystyle \prod_{r-1}^{n}\left ( \cos r\theta  +i\sin r\theta  \right )$$ equals


A
cos(n+12)θ+isin(n+12)θ
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B
cos(n(n1)2θ)+isin(n(n1)2)0
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C
inr=1(sinrθ+icosrθ)
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D
None of these
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Solution

The correct option is D None of these
$$z=\prod _{ r=1 }^{ n }{ \left( \cos { r\theta  } +i\sin { r\theta  }  \right)  } =\prod _{ r=1 }^{ n }{ { e }^{ ir\theta  } } =\prod _{ r=1 }^{ n }{ { \left( { e }^{ i\theta  } \right)  }^{ r } } $$
let $$a={ e }^{ i\theta  }$$
$$\Rightarrow z=\prod _{ r=1 }^{ n }{ { a }^{ r } } ={ a }^{ \sum _{ r=1 }^{ n }{ r }  }={ a }^{ \dfrac { n\left( n+1 \right)  }{ 2 }  }$$
$$\Rightarrow z=\left( { e }^{ i\theta  } \right) ^{ \dfrac { n\left( n+1 \right)  }{ 2 }  }={ e }^{ \dfrac { in\left( n+1 \right) \theta  }{ 2 }  }$$
$$\therefore \quad z=\cos { \left( \dfrac { n\left( n+1 \right) \theta  }{ 2 }  \right)  } +i\sin { \left( \dfrac { n\left( n+1 \right) \theta  }{ 2 }  \right)  } $$
Hence, option 'D' is correct.

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