<s>[INST] sum of perpendiculars of a regular 24 sided shape inscribed in a circle 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. 

 [/INST] 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 $r (   2 + \sqrt{2} + \sqrt{3} +\sqrt{6}   )$
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