DLMF:1.15.E15 (Q557)

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DLMF:1.15.E15
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    K n ( θ ) = 1 n + 1 ( sin ( 1 2 ( n + 1 ) θ ) sin ( 1 2 θ ) ) 2 , subscript 𝐾 𝑛 𝜃 1 𝑛 1 superscript 1 2 𝑛 1 𝜃 1 2 𝜃 2 {\displaystyle{\displaystyle K_{n}(\theta)=\frac{1}{n+1}\left(\frac{\sin\left(% \tfrac{1}{2}(n+1)\theta\right)}{\sin\left(\tfrac{1}{2}\theta\right)}\right)^{2% },}}
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    DLMF:1.15.E15
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