Board index Special Functions Simplify: sin(π/11)sin(2π/11)cos(π/22)cos(3π/22)cos(5π/22)

Simplify: sin(π/11)sin(2π/11)cos(π/22)cos(3π/22)cos(5π/22)

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Moderator: Shobhit



Posts: 50
Location: Redmond, WA

This post is not really about special functions, it's just about a trig expression. I found that it has a very simple form in radicals, but, curiously enough, it cannot be simplified directly to that form by Mathematica or Maple (although they can be manipulated to get to this form eventually through some directed transformations).

So, I present you this problem. Hope you will enjoy it.

Simplify:
$$\sin\left(\frac\pi{11}\right)\sin\left(\frac{2\pi}{11}\right)\cos\left(\frac\pi{22}\right)\cos\left(\frac{3\pi}{22}\right)\cos\left(\frac{5\pi}{22}\right)$$


Posts: 44
Let $S$ be the product. It can be simplified to:

$$S=\sin\left(\frac{\pi}{11}\right)\sin\left(\frac{2\pi}{11}\right)\sin\left(\frac{3\pi}{11}\right)\sin\left(\frac{4\pi}{11}\right)\sin\left(\frac{5\pi}{11}\right)=\sin\left(\frac{6\pi}{11}\right)\sin\left(\frac{7\pi}{11}\right)\sin\left(\frac{8\pi}{11}\right)\sin\left(\frac{9\pi}{11}\right)\sin\left(\frac{10\pi}{11}\right)$$


$$\Rightarrow S^2=\prod_{k=1}^{10}\sin\left(\frac{k\pi}{11}\right)$$

Since
$$\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$

$$\Rightarrow S^2=\frac{11}{2^{10}} \Rightarrow \boxed{S=\dfrac{\sqrt{11}}{32}}$$


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