sin 4 5 ∘ sin 4 6 ∘ sin 1 ∘ + sin 4 7 ∘ sin 4 8 ∘ sin 1 ∘ + sin 4 9 ∘ sin 5 0 ∘ sin 1 ∘ + ⋯ ⋯ + sin 1 3 1 ∘ sin 1 3 2 ∘ sin 1 ∘ + sin 1 3 3 ∘ sin 1 3 4 ∘ sin 1 ∘ = ?
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Exactly the same solution!!
This is from "103 Trigonometry Problems" by Titu Andreescu.
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S = sin 4 5 ∘ sin 4 6 ∘ sin 1 ∘ + sin 4 7 ∘ sin 4 8 ∘ sin 1 ∘ + sin 4 9 ∘ sin 5 0 ∘ sin 1 ∘ + . . . + sin 1 3 3 ∘ sin 1 3 4 ∘ sin 1 ∘ = n = 0 ∑ 4 4 sin ( 4 5 + 2 n ) ∘ sin ( 4 5 + 2 n + 1 ) ∘ sin 1 ∘ = n = 0 ∑ 4 4 sin ( 4 5 + 2 n ) ∘ sin ( 4 5 + 2 n + 1 ) ∘ sin ( 4 5 + 2 n + 1 − 4 5 + 2 n ) ∘ = n = 0 ∑ 4 4 sin ( 4 5 + 2 n ) ∘ sin ( 4 5 + 2 n + 1 ) ∘ sin ( 4 5 + 2 n + 1 ) ∘ cos ( 4 5 + 2 n ) ∘ − sin ( 4 5 + 2 n ) ∘ cos ( 4 5 + 2 n + 1 ) ∘ = n = 0 ∑ 4 4 ( cot ( 4 5 + 2 n ) ∘ − cot ( 4 5 + 2 n + 1 ) ∘ ) = n = 4 5 ∑ 1 3 4 ( − 1 ) n + 1 cot n ∘ = n = 4 5 ∑ 8 9 ( − 1 ) n + 1 cot n ∘ − cot 9 0 ∘ + n = 9 1 ∑ 1 3 4 ( − 1 ) n + 1 cot n ∘ As cot ( 1 8 0 ∘ − θ ) = − cot θ = n = 4 5 ∑ 8 9 ( − 1 ) n + 1 cot n ∘ − 0 + n = 9 1 ∑ 1 3 4 ( − 1 ) n cot ( 1 8 0 − n ) ∘ = n = 4 5 ∑ 8 9 ( − 1 ) n + 1 cot n ∘ + n = 4 6 ∑ 8 9 ( − 1 ) n cot n ∘ = cot 4 5 ∘ = 1