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12三角函数★★★☆☆
观察下列等式:
(sin⁡π3)−2+(sin⁡2π3)−2=43×1×2\left(\sin\dfrac{\pi}{3}\right)^{-2}+\left(\sin\dfrac{2\pi}{3}\right)^{-2}=\dfrac43\times1\times2;
(sin⁡π5)−2+(sin⁡2π5)−2+(sin⁡3π5)−2+(sin⁡4π5)−2=43×2×3\left(\sin\dfrac{\pi}{5}\right)^{-2}+\left(\sin\dfrac{2\pi}{5}\right)^{-2}+\left(\sin\dfrac{3\pi}{5}\right)^{-2}+\left(\sin\dfrac{4\pi}{5}\right)^{-2}=\dfrac43\times2\times3;
(sin⁡π7)−2+(sin⁡2π7)−2+(sin⁡3π7)−2+⋯+(sin⁡6π7)−2=43×3×4\left(\sin\dfrac{\pi}{7}\right)^{-2}+\left(\sin\dfrac{2\pi}{7}\right)^{-2}+\left(\sin\dfrac{3\pi}{7}\right)^{-2}+\cdots+\left(\sin\dfrac{6\pi}{7}\right)^{-2}=\dfrac43\times3\times4;
(sin⁡π9)−2+(sin⁡2π9)−2+(sin⁡3π9)−2+⋯+(sin⁡8π9)−2=43×4×5\left(\sin\dfrac{\pi}{9}\right)^{-2}+\left(\sin\dfrac{2\pi}{9}\right)^{-2}+\left(\sin\dfrac{3\pi}{9}\right)^{-2}+\cdots+\left(\sin\dfrac{8\pi}{9}\right)^{-2}=\dfrac43\times4\times5.
照此规律, (sin⁡π2n+1)−2+(sin⁡2π2n+1)−2+(sin⁡3π2n+1)−2+⋯+(sin⁡2nπ2n+1)−2=\left(\sin\dfrac{\pi}{2n+1}\right)^{-2}+\left(\sin\dfrac{2\pi}{2n+1}\right)^{-2}+\left(\sin\dfrac{3\pi}{2n+1}\right)^{-2}+\cdots+\left(\sin\dfrac{2n\pi}{2n+1}\right)^{-2}=______.

2016 · 山东 · 文 · 第 12 题 · 填空题

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