JEE Advanced 2016

Geometry Level 3

∑ k = 1 13 1 sin ⁡ ( π 4 + ( k − 1 ) π 6 ) sin ⁡ ( π 4 + k π 6 ) = ? \large\sum_{k = 1}^{13}\frac{1}{\sin\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right)\sin\left(\frac{\pi}{4} + \frac{k\pi}{6}\right)}= \, ?

2 + 3 2 +\sqrt3 2 ( 3 − 1 ) 2(\sqrt3 -1) 3 − 3 3 - \sqrt3 2 ( 3 − 3 ) 2(3-\sqrt3)

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Tapas Mazumdar
May 9, 2017

∑ k = 1 13 1 sin ⁡ ( π 4 + π 6 ( k − 1 ) ) sin ⁡ ( π 4 + π k 6 ) = ∑ k = 1 13 csc ⁡ π 6 ⋅ sin ⁡ π 6 sin ⁡ ( π 4 + π 6 ( k − 1 ) ) sin ⁡ ( π 4 + π k 6 ) = csc ⁡ π 6 ⋅ ∑ k = 1 13 sin ⁡ [ ( π 4 + π k 6 ) − ( π 4 + π 6 ( k − 1 ) ) ] sin ⁡ ( π 4 + π 6 ( k − 1 ) ) sin ⁡ ( π 4 + π k 6 ) = 2 ⋅ ∑ k = 1 13 sin ⁡ ( π 4 + π k 6 ) cos ⁡ ( π 4 + π 6 ( k − 1 ) ) − cos ⁡ ( π 4 + π k 6 ) sin ⁡ ( π 4 + π 6 ( k − 1 ) ) sin ⁡ ( π 4 + π 6 ( k − 1 ) ) sin ⁡ ( π 4 + π k 6 ) = 2 ⋅ ∑ k = 1 13 cot ⁡ ( π 4 + π 6 ( k − 1 ) ) − cot ⁡ ( π 4 + π k 6 ) = 2 [ cot ⁡ π 4 − cot ⁡ ( π 4 + 13 π 6 ) ] = 2 [ 1 − cot ⁡ 5 π 12 ] = 2 [ 1 − ( 2 − 3 ) ] = 2 ( 3 − 1 ) = 1.464 \begin{aligned} & \displaystyle \sum_{k=1}^{13} \dfrac{1}{\sin \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) \sin \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right)} \\ \\ &= \sum_{k=1}^{13} \csc \dfrac{\pi}{6} \cdot \dfrac{\sin \dfrac{\pi}{6}}{\sin \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) \sin \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right)} \\ \\ &= \displaystyle \csc \dfrac{\pi}{6} \cdot \sum_{k=1}^{13} \dfrac{\sin \left[ \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right) - \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) \right] }{ \sin \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) \sin \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right)} \\ \\ &= \displaystyle 2 \cdot \sum_{k=1}^{13} \dfrac{ \sin \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right) \cos \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) - \cos \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right) \sin \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) }{ \sin \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) \sin \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right)} \\ \\ &= \displaystyle 2 \cdot \sum_{k=1}^{13} \cot \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} (k-1) \right) - \cot \left( \dfrac{\pi}{4} + \dfrac{\pi k}{6} \right) \\ \\ &= 2 \left[ \cot \dfrac{\pi}{4} - \cot \left( \dfrac{\pi}{4} + \dfrac{13 \pi}{6} \right) \right] \\ \\ &= 2 \left[ 1 - \cot \dfrac{5 \pi}{12} \right] \\ \\ &= 2 \left[ 1 - (2 - \sqrt{3}) \right] \\ \\ &= 2 ( \sqrt{3} - 1 ) \\ \\ &= \boxed{1.464} \end{aligned}

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Tapas Mazumdar - 4 years, 1 month ago

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