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The following question is from Gelfand and Saul's book 'Trigonometry'

The question follows from a section about summing trigonometric series using a 'telescoping sum' method. I guess this is the method the authors intend.

I apologise for the apparent laziness in posting an image of the question. I have thought about this a lot and have no idea where to begin. It is not a homework question, but is something I have come across during self study.

enter image description here

mikoyan
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    It might help to express the desired quantity in summation notation. Connecting the (upper) vertices of the $24$-gon to the circle's center, each perpendicular is the height of a right triangle with hypotenuse equal to the circle's radius (call it $r$); the angles at the center are clearly multiples of $\pi/12$. So, you're looking for $$r\sin\frac{\pi}{12} + r\sin\frac{2\pi}{12} + \cdots + r\sin\frac{11\pi}{12} = r;\sum_{k=0}^{12} \sin\frac{k\pi}{12}$$(where I've thrown in the cases $k=0$ and $k=12$, which contribute nothing to the sum, for the sake of the pattern). Can you go from here? – Blue Nov 07 '13 at 22:06
  • @Blue That was very helpful, thank you. On calculating the series and letting $r=1$ I got the answer $2+\sqrt{2} + \sqrt{3} + \sqrt{6}$ so for any value of r the answer would be: $r( 2+\sqrt{2} + \sqrt{3} + \sqrt{6})$ ) – mikoyan Nov 08 '13 at 19:50

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Well Blue provided the answer with the series, which written out in 'longhand' and letting $r=1$is:

$$\sin \frac{0\pi}{12} = 0$$ $$\sin \frac{1\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$$ $$\sin \frac{2\pi}{12} = \frac{1}{2}$$ $$\sin \frac{3\pi}{12} = \frac{\sqrt{2}}{2}$$ $$\sin \frac{4\pi}{12} = \frac{\sqrt{3}}{2}$$ $$\sin \frac{5\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$$ $$\sin \frac{6\pi}{12} = 1$$ $$\sin \frac{7\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$$ $$\sin \frac{8\pi}{12} = \frac{\sqrt{3}}{2}$$ $$\sin \frac{9\pi}{12} = \frac{\sqrt{2}}{2}$$ $$\sin \frac{10\pi}{12} = \frac{1}{2}$$ $$\sin \frac{11\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$$ $$\sin \frac{12\pi}{12} = 0$$

Which sums to $2 + \sqrt{2} + \sqrt{3} +\sqrt{6}$ and for any value of $r$ is $r ( 2 + \sqrt{2} + \sqrt{3} +\sqrt{6} )$

mikoyan
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  • You could simplify your answer greatly. By noting that $\sin(\pi-\theta)=\sin\theta$. So you could sum till $\frac{5\pi}{12}$, double that and add 1. – Sawarnik Feb 24 '14 at 04:18
  • Second point, you could use the identity: $$\sin a + \sin 3a + \sin 5a = \frac{(\sin 3a)^2}{\sin a}$$ So: $$\sin \frac{\pi}{12} + \sin \frac{3\pi}{12} + \sin \frac{5\pi}{12} = \frac{1}{2\sin \frac{\pi}{12}}$$ – Sawarnik Feb 24 '14 at 04:24
  • $$\frac{1}{2\sin \frac{\pi}{12}} = \frac{2}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{2}$$ Now just $\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$. To get: $$\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+1}{2}$$
    Now double and add 1.     – Sawarnik Feb 24 '14 at 04:31
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Hint: Try starting with a hexagon inscribed in a circle with radius r. Draw lines from the center to each vertex. This will make six equilateral triangles. The perpendiculars will form 30-60-90 triangles. For each right triangle you will have have a length r and all three angles. Can you calculate the lengths of the other sides of the triangle?

Steve ODonnell
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