k = 1 ∑ 1 3 sin ( 4 π + 6 ( k − 1 ) π ) sin ( 4 π + 6 k π ) 1 = ? Given the answer in 3 decimal places.
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Very nice solution(+1). Did the same way
@Tapas Mazumdar very nice solution and i like the way you start with the problem. :)
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k = 1 ∑ 1 3 sin ( 4 π + 6 π ( k − 1 ) ) sin ( 4 π + 6 π k ) 1 = k = 1 ∑ 1 3 csc 6 π ⋅ sin ( 4 π + 6 π ( k − 1 ) ) sin ( 4 π + 6 π k ) sin 6 π = csc 6 π ⋅ k = 1 ∑ 1 3 sin ( 4 π + 6 π ( k − 1 ) ) sin ( 4 π + 6 π k ) sin [ ( 4 π + 6 π k ) − ( 4 π + 6 π ( k − 1 ) ) ] = 2 ⋅ k = 1 ∑ 1 3 sin ( 4 π + 6 π ( k − 1 ) ) sin ( 4 π + 6 π k ) sin ( 4 π + 6 π k ) cos ( 4 π + 6 π ( k − 1 ) ) − cos ( 4 π + 6 π k ) sin ( 4 π + 6 π ( k − 1 ) ) = 2 ⋅ k = 1 ∑ 1 3 cot ( 4 π + 6 π ( k − 1 ) ) − cot ( 4 π + 6 π k ) = 2 [ cot 4 π − cot ( 4 π + 6 1 3 π ) ] = 2 [ 1 − cot 1 2 5 π ] = 2 [ 1 − ( 2 − 3 ) ] = 2 ( 3 − 1 ) = 1 . 4 6 4