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Question Number 152265 by mathdanisur last updated on 26/Aug/21
∑nk=1sin−1(k−k−1k(k+1)=?
Answered by mindispower last updated on 27/Aug/21
sin−(a)−sin−(b)=sin−(a1−b2−b1−a2)k−k−1k(k+1)=1k+1−k−1k.k+1−=1k1−(1k+1)2−1k+11−(1k)2=1k+1−k−1k.k+1⇒sin−(k−k−1k(k+1)=sin−(1k)−sin−(1k+1)⇔∑nk=1sin−(k−k−1k(k+1))=sin−(1n)−π4
Commented by mathdanisur last updated on 27/Aug/21
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