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Is it possible get an algebraic expression for $sin 1°$ that does not contain roots of negative values?
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Is it possible get an algebraic expression for $sin 1°$ that does not contain roots of negative values, so that it can be evaluated entirely using only real numbers?
For example, this paper gives some algebraic expressions for it, but they all involve complex numbers.
trigonometry algebraic-number-theory real-numbers
edited Jan 27 at 20:02
Blue
49.2k870157
asked Jan 27 at 19:21
tmlentmlen
20617
$endgroup$
add a comment |
$begingroup$
Is it possible get an algebraic expression for $sin 1°$ that does not contain roots of negative values, so that it can be evaluated entirely using only real numbers?
For example, this paper gives some algebraic expressions for it, but they all involve complex numbers.
trigonometry algebraic-number-theory real-numbers
edited Jan 27 at 20:02
Blue
49.2k870157
asked Jan 27 at 19:21
tmlentmlen
20617
$endgroup$
$begingroup$
It seems to be hard dogding the usage of complex numbers here since, as far as I can tell from the given sources, it appears to be common sense reducing the whole problem back to $sin(3x)=3sin(x)-4sin^3(x)$ which leaves us with a cubic equation in $sin(x)$ $($assuming $sin(3)$ as known$)$. Solving this invokes complex numbers in a natural way due the general solving formulae for cubic equations.
$endgroup$
– mrtaurho
Jan 27 at 19:46
add a comment |
$begingroup$
Is it possible get an algebraic expression for $sin 1°$ that does not contain roots of negative values, so that it can be evaluated entirely using only real numbers?
For example, this paper gives some algebraic expressions for it, but they all involve complex numbers.
trigonometry algebraic-number-theory real-numbers
edited Jan 27 at 20:02
Blue
49.2k870157
asked Jan 27 at 19:21
tmlentmlen
20617
$endgroup$
Is it possible get an algebraic expression for $sin 1°$ that does not contain roots of negative values, so that it can be evaluated entirely using only real numbers?
For example, this paper gives some algebraic expressions for it, but they all involve complex numbers.
trigonometry algebraic-number-theory real-numbers
trigonometry algebraic-number-theory real-numbers
edited Jan 27 at 20:02
Blue
49.2k870157
asked Jan 27 at 19:21
tmlentmlen
20617
edited Jan 27 at 20:02
Blue
49.2k870157
asked Jan 27 at 19:21
tmlentmlen
20617
edited Jan 27 at 20:02
Blue
49.2k870157
edited Jan 27 at 20:02
Blue
49.2k870157
edited Jan 27 at 20:02
Blue
49.2k870157
49.2k870157
asked Jan 27 at 19:21
tmlentmlen
20617
asked Jan 27 at 19:21
tmlentmlen
20617
asked Jan 27 at 19:21
tmlentmlen
20617
20617
$begingroup$
It seems to be hard dogding the usage of complex numbers here since, as far as I can tell from the given sources, it appears to be common sense reducing the whole problem back to $sin(3x)=3sin(x)-4sin^3(x)$ which leaves us with a cubic equation in $sin(x)$ $($assuming $sin(3)$ as known$)$. Solving this invokes complex numbers in a natural way due the general solving formulae for cubic equations.
$endgroup$
– mrtaurho
Jan 27 at 19:46
add a comment |
$begingroup$
It seems to be hard dogding the usage of complex numbers here since, as far as I can tell from the given sources, it appears to be common sense reducing the whole problem back to $sin(3x)=3sin(x)-4sin^3(x)$ which leaves us with a cubic equation in $sin(x)$ $($assuming $sin(3)$ as known$)$. Solving this invokes complex numbers in a natural way due the general solving formulae for cubic equations.
$endgroup$
– mrtaurho
Jan 27 at 19:46
$begingroup$
It seems to be hard dogding the usage of complex numbers here since, as far as I can tell from the given sources, it appears to be common sense reducing the whole problem back to $sin(3x)=3sin(x)-4sin^3(x)$ which leaves us with a cubic equation in $sin(x)$ $($assuming $sin(3)$ as known$)$. Solving this invokes complex numbers in a natural way due the general solving formulae for cubic equations.
$endgroup$
– mrtaurho
Jan 27 at 19:46
$begingroup$
It seems to be hard dogding the usage of complex numbers here since, as far as I can tell from the given sources, it appears to be common sense reducing the whole problem back to $sin(3x)=3sin(x)-4sin^3(x)$ which leaves us with a cubic equation in $sin(x)$ $($assuming $sin(3)$ as known$)$. Solving this invokes complex numbers in a natural way due the general solving formulae for cubic equations.
$endgroup$
– mrtaurho
Jan 27 at 19:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If this would be possible, you could also get an algebraic expression for $sin(10^o)$ that doesn't contain complex numbers. That is one of three real roots
of the cubic $8 z^3 - 6 z + 1$. Now see casus irreducibilis: given a cubic with rational coefficients that is irreducible over the rationals and has three real roots, it is impossible to express the roots using real radicals.
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If this would be possible, you could also get an algebraic expression for $sin(10^o)$ that doesn't contain complex numbers. That is one of three real roots
of the cubic $8 z^3 - 6 z + 1$. Now see casus irreducibilis: given a cubic with rational coefficients that is irreducible over the rationals and has three real roots, it is impossible to express the roots using real radicals.
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
add a comment |
$begingroup$
If this would be possible, you could also get an algebraic expression for $sin(10^o)$ that doesn't contain complex numbers. That is one of three real roots
of the cubic $8 z^3 - 6 z + 1$. Now see casus irreducibilis: given a cubic with rational coefficients that is irreducible over the rationals and has three real roots, it is impossible to express the roots using real radicals.
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
add a comment |
$begingroup$
If this would be possible, you could also get an algebraic expression for $sin(10^o)$ that doesn't contain complex numbers. That is one of three real roots
of the cubic $8 z^3 - 6 z + 1$. Now see casus irreducibilis: given a cubic with rational coefficients that is irreducible over the rationals and has three real roots, it is impossible to express the roots using real radicals.
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
$endgroup$
If this would be possible, you could also get an algebraic expression for $sin(10^o)$ that doesn't contain complex numbers. That is one of three real roots
of the cubic $8 z^3 - 6 z + 1$. Now see casus irreducibilis: given a cubic with rational coefficients that is irreducible over the rationals and has three real roots, it is impossible to express the roots using real radicals.
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
answered Jan 27 at 19:59
Robert IsraelRobert Israel
329k23217470
329k23217470
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
add a comment |
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
$begingroup$
More generally, trig values (when defined) of rational degree angles can be expressed in terms of real radicals if and only if the values are constructible real numbers, a result that Andrzej W. Mostowski proved in this 1948 paper, although the result was known prior to this, possibly by Mostowski himself, although he doesn't say either way in the paper. (This paper appeared in the first volume of Colloquium Mathematicum, and the early volume of this journal often published student-authored and student-level non-research papers.)
$endgroup$
– Dave L. Renfro
Jan 27 at 20:19
add a comment |
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$begingroup$
It seems to be hard dogding the usage of complex numbers here since, as far as I can tell from the given sources, it appears to be common sense reducing the whole problem back to $sin(3x)=3sin(x)-4sin^3(x)$ which leaves us with a cubic equation in $sin(x)$ $($assuming $sin(3)$ as known$)$. Solving this invokes complex numbers in a natural way due the general solving formulae for cubic equations.
$endgroup$
– mrtaurho
Jan 27 at 19:46