Best solutions to - "I would like to show thatsin⁡π3sin⁡2π3⋅sin⁡3π13⋅…⋅sin⁡6π13=1364"

Camila Glenn

Answered question

2022-04-08

I would like to show that

\sin \frac{π}{3} \sin \frac{2 π}{3} \cdot \sin \frac{3 π}{13} \cdot \dots \cdot \sin \frac{6 π}{13} = \frac{\sqrt{13}}{64}

Videoad3u

Beginner2022-04-09Added 15 answers

Use this formula (found here, and mentioned recently on MSE here):

\prod_{k = 1}^{n - 1} \sin (\frac{k π}{n}) = \frac{n}{2^{n - 1}}

Let

n = 13

, which gives

(\sin \frac{π}{13} \cdot \frac{2 π}{13} \cdot \sin \frac{3 π}{13} \cdot \dots \cdot \frac{6 π}{13}) (\sin \frac{7 π}{13} \cdot \frac{8 π}{13} \cdot \sin \frac{9 π}{13} \dots \sin \frac{12 π}{13}) = \frac{13}{2^{12}}

Use the fact that

\sin \frac{k π}{13} = \sin \frac{(13 - k) π}{13}

to see that this is the same as

{(\sin \frac{π}{13} \cdot \sin \frac{2 π}{13} \cdot \sin \frac{3 π}{13} \dots \sin \frac{6 π}{13})}^{2} = \frac{13}{2^{12}}

and take the square root of both sides to get your answer.

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