jacobi1
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Find the exact value of $$\sin \frac{\pi}{180}$$.
The discussion centers around finding the exact value of the trigonometric function $$\sin \frac{\pi}{180}$$, which is equivalent to $$\sin 1^\circ$$. Participants explore various methods and formulas related to this calculation, including references to cubic equations and known sine values.
Participants express uncertainty regarding the argument's units (radians vs. degrees). There are multiple approaches to finding the exact value of $$\sin 1^\circ$$, and no consensus is reached on a definitive method or solution.
Some methods rely on specific formulas and assumptions about the cubic equation derived from known sine values, which may not be universally accepted or understood.
Ummmmm...is the argument in radians or degrees? Kinda hard to tell given the fraction.jacobi said:Find the exact value of $$\sin \frac{\pi}{180}$$.
Start with the known formula for $\sin 3^\circ$ (you can find it here): $\sin 3^\circ = \frac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr).$ The formula $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$ then tells you that $\sin1^\circ$ is the smaller of the two positive roots of the cubic equation $$4x^3 - 3x + \tfrac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr) = 0,$$ which can be solved exactly, for example by Vieta's method. But don't expect a neat solution. (Emo)jacobi said:Find the exact value of $$\sin \frac{\pi}{180}$$ (in other words, $\color{red}{\sin 1^\circ}$).
By convention, if there is no degree symbol in the argument, it is in radians.topsquark said:Ummmmm...is the argument in radians or degrees? Kinda hard to tell given the fraction.
-Dan
Opalg said:Start with the known formula for $\sin 3^\circ$ (you can find it here): $\sin 3^\circ = \frac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr).$ The formula $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$ then tells you that $\sin1^\circ$ is the smaller of the two positive roots of the cubic equation $$4x^3 - 3x + \tfrac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr) = 0,$$ which can be solved exactly, for example by Vieta's method. But don't expect a neat solution. (Emo)