The value of ∑ k = 1 13 1 sin ⁡ π 4 + k - 1 π 6 sin ⁡ π 4 + k π 6 is…

The value of ∑k=1131sin⁡π4+k-1π6sin⁡π4+kπ6 is equal to
  1. 3- 3
  2. 23- 3
  3. 23-1
  4. 22+ 3

Solution

Let π 4 +( K−1 ) π 6 =θ
⇒ ∑ k=1 13 1 sinθsin( θ+ π 6 ) = 1 sin π 6 ∑ k=1 13 sin π 6 sinθsin( θ+ π 6 )
=2 ∑ k=1 13 sin( θ+ π 6 −θ ) sinθsin( θ+ π 6 ) =2 ∑ k=1 13 sin( θ+ π 6 )cosθ−cos( θ+ π 6 )sinθ sinθsin( θ+ π 6 )
=2 ∑ k=1 13 ( cotθ−cot( θ+ π 6 ) )=2 ∑ k=1 13 ( cot( π 4 +( k−1 ) π 6 )−cot( π 4 + kπ 6 ) )
=2( cot π 4 −cot( π 4 + π 6 )+cot( π 4 + π 6 )−cot( π 4 + 2π 6 )+−−−−−cot( π 4 + 12π 6 )−cot( π 4 + 13π 6 ) )
=2[ cot π 4 −cot( π 4 + 13π 6 ) ]=2[ 1−cot 29π 12 ]=2[ 1−cot 75 ο ]=2[ 1−( 2− 3 ) ]=2( 3−1 )

Asked in: JEE Advanced 2016 (Paper 2)

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