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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
where all angles are in degree
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 definitively calculated as \(4.0194366942304562 \times 10^{-14}\). This calculation was discussed among forum members, with one user, Sudhakara, noting the difficulty in solving it algebraically. Alternative methods, including the use of complex numbers and nth roots of unity, were suggested for further exploration.
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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
Sudharaka said:Hi jacks,
I haven't found a way solve this algebraically. But if you are interested about the answer it is, \(4.0194366942304562\times 10^{-14}\)
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 $\cos(90x) = 0.$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