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Geometry Level 4

r = 0 502 ( 1 ) r sin ( ( 2 r + 1 ) π 2014 ) = ? \large \sum_{r=0}^{502} (-1)^r \sin\left( \dfrac{(2r+1)\pi}{2014} \right) = \, ?


The answer is 0.5.

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1 solution

Abhishek Sharma
Jan 20, 2016

r = 0 502 ( 1 ) r sin ( π 2014 ( 2 r + 1 ) ) \sum _{ r=0 }^{ 502 }{ { \left( -1 \right) }^{ r }\sin { \left( \frac { \pi }{ 2014 } \left( 2r+1 \right) \right) } } 1 2 cos π 2014 r = 0 502 ( 1 ) r [ 2 sin ( π 2014 ( 2 r + 1 ) ) cos π 2014 ] \frac { 1 }{ 2\cos { \frac { \pi }{ 2014 } } } \sum _{ r=0 }^{ 502 }{ { \left( -1 \right) }^{ r }\left[ 2\sin { \left( \frac { \pi }{ 2014 } \left( 2r+1 \right) \right) \cos { \frac { \pi }{ 2014 } } } \right] } 1 2 cos π 2014 r = 0 502 ( 1 ) r [ sin ( π 1007 ( r + 1 ) ) + sin ( π 1007 ( r ) ) ] \frac { 1 }{ 2\cos { \frac { \pi }{ 2014 } } } \sum _{ r=0 }^{ 502 }{ { \left( -1 \right) }^{ r }\left[ \sin { \left( \frac { \pi }{ 1007 } \left( r+1 \right) \right) } +\sin { \left( \frac { \pi }{ 1007 } \left( r \right) \right) } \right] } 1 2 cos π 2014 r = 0 502 ( 1 ) r sin ( π 1007 ( r ) ) + ( 1 ) r sin ( π 1007 ( r + 1 ) ) \frac { 1 }{ 2\cos { \frac { \pi }{ 2014 } } } \sum _{ r=0 }^{ 502 }{ { \left( -1 \right) }^{ r }\sin { \left( \frac { \pi }{ 1007 } \left( r \right) \right) } +{ \left( -1 \right) }^{ r }\sin { \left( \frac { \pi }{ 1007 } \left( r+1 \right) \right) } } 1 2 cos π 2014 r = 0 502 ( 1 ) r sin ( π 1007 ( r ) ) ( 1 ) r + 1 sin ( π 1007 ( r + 1 ) ) \frac { 1 }{ 2\cos { \frac { \pi }{ 2014 } } } \sum _{ r=0 }^{ 502 }{ { \left( -1 \right) }^{ r }\sin { \left( \frac { \pi }{ 1007 } \left( r \right) \right) } -{ \left( -1 \right) }^{ r+1 }\sin { \left( \frac { \pi }{ 1007 } \left( r+1 \right) \right) } } 1 2 cos π 2014 ( ( 1 ) 0 sin ( π 1007 ( 0 ) ) ( 1 ) 502 + 1 sin ( π 1007 ( 502 + 1 ) ) ) \frac { 1 }{ 2\cos { \frac { \pi }{ 2014 } } } \left( { \left( -1 \right) }^{ 0 }\sin { \left( \frac { \pi }{ 1007 } \left( 0 \right) \right) } -{ \left( -1 \right) }^{ 502+1 }\sin { \left( \frac { \pi }{ 1007 } \left( 502+1 \right) \right) } \right) sin 503 π 1007 2 cos π 2014 \frac { \sin { \frac { 503\pi }{ 1007 } } }{ 2\cos { \frac { \pi }{ 2014 } } } cos ( π 2 503 π 1007 ) 2 cos π 2014 \frac { \cos { ( \frac { \pi }{ 2 } -\frac { 503\pi }{ 1007 } ) } }{ 2\cos { \frac { \pi }{ 2014 } } } 1 2 \frac { 1 }{ 2 }

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