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

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.Maths

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