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I have the exact values for the sine of integers. Has this been accomplished before? Jim Parent

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And by "exact value", what do you mean? Most values are irrational, so they have a nonterminating, nonrepeating decimal expansion. How are you expressing these values? –  Arturo Magidin Jun 10 '11 at 20:46
    
Integer number of degrees? Do you allow cube roots of non-real complex numbers in your expression? –  André Nicolas Jun 10 '11 at 21:04
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I would ask you to explain yourself further, and moreover, I would advise you to temper your enthusiasm... if you think you have an exact value of, say $\sin{1}$, I would hesitate to believe you, as this is, as far as I know, a known transcendental number for which the simplest expression is quite simply $\sin{1}...$ –  barf Jun 10 '11 at 21:56
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@JTL, I expect James is working in degrees, not radians. –  Gerry Myerson Jun 10 '11 at 23:29
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The exact value of $sin(1°)$ is derived here: efnet-math.org/Meta/sine1.htm PS: I didn't check it ! –  tos Jan 31 '12 at 23:21
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In radians, the sine of any nonzero integer angle is a transcendental number. This follows from Lindemann–Weierstrass theorem. As far as I know, there is no closed-form expression for $\sin(1)$ that doesn't somehow involve transcendental functions

Using degrees, the sine of any integer is an algebraic number, i.e. a root of a polynomial with integer coefficients. For $\sin(1^\circ)$, the smallest such polynomial has degree 48. It is also possible to express $\sin(1^\circ)$ by radicals as follows: $$ \sin(1^\circ) \;=\; \frac{(\sqrt[180]{-1})^{89} - (\sqrt[180]{-1})^{91}}{2} $$ where $\sqrt[180]{-1}$ denotes the principal 180'th root of $-1$, i.e. $\sqrt[180]{-1} = \cos(1^\circ) + i \sin(1^\circ)$. More generally, $$ \sin(k^\circ) \;=\; \frac{(\sqrt[180]{-1})^{90-k} - (\sqrt[180]{-1})^{90+k}}{2} $$ for any integer $k$.

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