Solve $x\sqrt{10} = \prod\limits_{k = 1}^{90} \sin(k), x\in \mathbb Q$.

Question

Can someone help me with this question? I've found a solution but it's not a very nice one. I used 6 times the relation $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. There's got to be a better way.

$x\sqrt{10} = \prod\limits_{k = 1}^{90} \sin(k)$, k in degrees.

Presumably $k$ is in degrees, and the product is a rational multiple of $\sqrt{10}$, which Pedro was assigned to find. — GEdgar, Jul 04 '13 at 22:04

score 8 · Accepted Answer · edited Apr 13 '17 at 12:19

8

We use this, which is somewhat complicated.

Let $S$ be the product $\prod_{k=1}^{90} \sin k^\circ $. Then $S^2 = \prod_{k=1}^{179} \sin k^\circ = \frac{ 180} { 2^{179}}$

Hence $S = \sqrt{ 10} \frac{3}{2^{89} }$.

I believe your method of using $\sin 2\theta$ repeatedly is better, in part because the proof of the quoted theorem is complicated.

edited Apr 13 '17 at 12:19

Community

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answered Jul 04 '13 at 22:32

Calvin Lin

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Solve $x\sqrt{10} = \prod\limits_{k = 1}^{90} \sin(k), x\in \mathbb Q$.

1 Answers1