A geometry problem by Rahil Sehgal

Geometry Level 4

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

Find the value of the expression to about 3 decimal places.


The answer is 11.758.

Rahil Sehgal by Rahil Sehgal , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tapas Mazumdar
May 1, 2017

∑ k = 1 13 sin ⁡ ( π k 6 + π 4 ) sin ⁡ ( π 6 ( k − 1 ) + π 4 ) = ∑ k = 1 13 sin ⁡ ( ( π 6 ( k − 1 ) + π 4 ) + π 6 ) sin ⁡ ( π 6 ( k − 1 ) + π 4 ) = ∑ k = 1 13 sin ⁡ ( π 6 ( k − 1 ) + π 4 ) cos ⁡ π 6 + cos ⁡ ( π 6 ( k − 1 ) + π 4 ) sin ⁡ π 6 sin ⁡ ( π 6 ( k − 1 ) + π 4 ) = ∑ k = 1 13 cos ⁡ π 6 + sin ⁡ π 6 ⋅ cot ⁡ ( π 6 ( k − 1 ) + π 4 ) = 13 cos ⁡ π 6 + sin ⁡ π 6 ∑ k = 1 13 cot ⁡ ( π 6 ( k − 1 ) + π 4 ) = 13 cos ⁡ π 6 + sin ⁡ π 6 ∑ k = 1 13 cot ⁡ ( π k 6 + π 12 ) \begin{aligned} \displaystyle \sum_{k=1}^{13} \dfrac{\sin \left( \dfrac{\pi k}{6} + \dfrac{\pi}{4} \right)}{\sin \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right)} &= \sum_{k=1}^{13} \dfrac{\sin \left( \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right) + \dfrac{\pi}{6} \right)}{\sin \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right)} \\ &= \displaystyle \sum_{k=1}^{13} \dfrac{ \sin \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right) \cos \dfrac{\pi}{6} + \cos \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right) \sin \dfrac{\pi}{6} }{ \sin \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right) } \\ &= \displaystyle \sum_{k=1}^{13} \cos \dfrac{\pi}{6} + \sin \dfrac{\pi}{6} \cdot \cot \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right) \\ &= \displaystyle 13 \cos \dfrac{\pi}{6} + \sin \dfrac{\pi}{6} \sum_{k=1}^{13} \cot \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right) \\ &= \displaystyle 13 \cos \dfrac{\pi}{6} + \sin \dfrac{\pi}{6} \sum_{k=1}^{13} \cot \left( \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right) \end{aligned}

We note that

cot ⁡ ( π k 6 + π 12 ) = − cot ⁡ ( π + π k 6 + π 12 ) ⟹ cot ⁡ ( π k 6 + π 12 ) = − cot ⁡ ( π 6 ( k + 6 ) + π 12 ) ⟹ cot ⁡ ( π k 6 + π 12 ) + cot ⁡ ( π 6 ( k + 6 ) + π 12 ) = 0 \begin{aligned} & \cot \left( \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right) = - \cot \left( \pi + \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right) \\ \implies & \cot \left( \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right) = - \cot \left( \dfrac{\pi}{6} (k+6) + \dfrac{\pi}{12} \right) \\ \implies & \cot \left( \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right) + \cot \left( \dfrac{\pi}{6} (k+6) + \dfrac{\pi}{12} \right) = 0 \end{aligned}

Therefore

∑ k = 1 13 cot ⁡ ( π k 6 + π 12 ) = ∑ k = 1 12 cot ⁡ ( π k 6 + π 12 ) ⏟ = 0 + cot ⁡ ( 13 π 6 + π 12 ) = cot ⁡ 27 π 12 = cot ⁡ 9 π 4 = 1 \displaystyle \sum_{k=1}^{13} \cot \left( \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right) = \underbrace{\sum_{k=1}^{12} \cot \left( \dfrac{\pi k}{6} + \dfrac{\pi}{12} \right)}_{\color{#3D99F6} = \ 0} + \cot \left( \dfrac{13 \pi}{6} + \dfrac{\pi}{12} \right) = \cot \dfrac{27 \pi}{12} = \cot \dfrac{9 \pi}{4} = 1

Hence we get

∑ k = 1 13 sin ⁡ ( π k 6 + π 4 ) sin ⁡ ( π 6 ( k − 1 ) + π 4 ) = 13 cos ⁡ π 6 + sin ⁡ π 6 = 13 3 + 1 2 = 11.758 \displaystyle \sum_{k=1}^{13} \dfrac{\sin \left( \dfrac{\pi k}{6} + \dfrac{\pi}{4} \right)}{\sin \left( \dfrac{\pi}{6} (k-1) + \dfrac{\pi}{4} \right)} = 13 \cos \dfrac{\pi}{6} + \sin \dfrac{\pi}{6} = \dfrac{13 \sqrt{3} +1}{2} = \boxed{11.758}

6 Helpful 1 Interesting 1 Brilliant 0 Confused

Very Good solution. (+1)

Rahil Sehgal - 4 years, 1 month ago

Log in to reply

Thank you. :)

Tapas Mazumdar - 4 years, 1 month ago

Did the same as your approach.(+1)

Aditya Kumar - 4 years ago

Log in to reply

Great job! :D

Tapas Mazumdar - 4 years ago

@Tapas Mazumdar Help me out with the step where you wrote 'We note that'.

Ankit Kumar Jain - 3 years ago

Log in to reply

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...