MHB Sin Values of 87 and 89 Degrees

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The value of the product $\sin(1^\circ) \cdot \sin(3^\circ) \cdot \sin(5^\circ) \cdots \sin(87^\circ) \cdot \sin(89^\circ)$ is approximately \(4.0194366942304562 \times 10^{-14}\). Attempts to solve this algebraically have been unsuccessful, with some participants exploring complex numbers and nth roots of unity for potential solutions. A reference method involving cosine functions has been suggested to approach the problem. The discussion highlights the challenge of finding a straightforward algebraic solution. Overall, the focus remains on determining the value of the sine product.
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The value of $\sin (1^0).\sin (3^0).\sin (5^0)...\sin (87^0).\sin (89^0)$

where all angles are in degree
 
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jacks said:
The value of $\sin (1^0).\sin (3^0).\sin (5^0)...\sin (87^0).\sin (89^0)$

where all angles are in degree

Hi jacks,

I haven't found a way solve this algebraically. But if you are interested about the answer it is, \(4.0194366942304562\times 10^{-14}\)
 
Sudharaka said:
Hi jacks,

I haven't found a way solve this algebraically. But if you are interested about the answer it is, \(4.0194366942304562\times 10^{-14}\)

Thanks Sudhakara

I am trying to find it with the help of complex no.(like nth -roots of unity)
 
jacks said:
The value of $\sin (1^0).\sin (3^0).\sin (5^0)...\sin (87^0).\sin (89^0)$

where all angles are in degree
Follow the method used in http://www.mathhelpboards.com/showthread.php?253-Simplify-cos(a)cos(2a)cos(3a)-cos(999a)-if-a-(2pi)-1999&p=1517&viewfull=1#post1517, noting that $x=\pm1^\circ,\pm3^\circ,\pm5^\circ,\ldots,\pm89 ^\circ$ are the solutions of the equation $\cos(90x) = 0.$
 

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